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<channel>
	<title>substitusi &amp;laquo; WordPress.com Tag Feed</title>
	<link>http://en.wordpress.com/tag/substitusi/</link>
	<description>Feed of posts on WordPress.com tagged "substitusi"</description>
	<pubDate>Wed, 10 Feb 2010 15:02:41 +0000</pubDate>

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	<language>en</language>

<item>
<title><![CDATA[(Bag.3) Semakin Hebat Eliminasi Sistem Persamaan Linear Aljabar]]></title>
<link>http://apiqquantum.wordpress.com/2009/10/06/bag-3-semakin-hebat-eliminasi-sistem-persamaan-linear-aljabar/</link>
<pubDate>Tue, 06 Oct 2009 22:59:13 +0000</pubDate>
<dc:creator>apiqquantum</dc:creator>
<guid>http://apiqquantum.wordpress.com/2009/10/06/bag-3-semakin-hebat-eliminasi-sistem-persamaan-linear-aljabar/</guid>
<description><![CDATA[&#8220;Ada apa Al?&#8221; tanya Paman APIQ. &#8220;Aku bingung nih&#8230;&#8221; &#8220;Bersyukurlah]]></description>
<content:encoded><![CDATA[&#8220;Ada apa Al?&#8221; tanya Paman APIQ. &#8220;Aku bingung nih&#8230;&#8221; &#8220;Bersyukurlah]]></content:encoded>
</item>
<item>
<title><![CDATA[DI BALIK ANOMALI HARGA GULA]]></title>
<link>http://hagemman.wordpress.com/2009/09/07/di-balik-anomali-harga-gula/</link>
<pubDate>Mon, 07 Sep 2009 02:44:22 +0000</pubDate>
<dc:creator>hagemman</dc:creator>
<guid>http://hagemman.wordpress.com/2009/09/07/di-balik-anomali-harga-gula/</guid>
<description><![CDATA[Terasa janggal dan penuh kontradiksi ketika berlangsung panen raya tebu atau giling pabrik gula, har]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><img class="alignleft size-thumbnail wp-image-2455" title="sugarcane" src="http://hagemman.wordpress.com/files/2009/09/sugarcane.jpg?w=129" alt="sugarcane" width="129" height="150" />Terasa janggal dan penuh kontradiksi ketika berlangsung panen raya tebu atau giling pabrik gula, harga gula melambung tinggi, apalagi untuk Jatim yang notabene merupakan sentra produksi gula terpenting. Dengan keberadaan 31 unit PG, pada tahun 2009 ini Jatim diperkirakan mampu memasok 1,3 juta ton daro 2,7 juta ton produksi gula secara nasional. Dengan kebutuhan hanya sekitar 480.000 – 550.000 ton, terjadi surplus yang selama ini dialirkan di propinsi lain yang bukan merupakan produsen.</p>
<p>Anomali harga terjadi sejak harga gula diserahkan kepada mekanisme pasar. Pengertiannya, saat jumlah barang ditawarkan di pasar jauh lebih banyak dibanding permintaan, harga pasti terjungkal. Sebaliknya, begitu jumlah barang ditawarkan lebih sedikit ketimbang permintaan, meroketnya harga tak dapat dihindari.</p>
<p>Hukum ekonomi tadi juga berlaku mutlak untuk gula, khususnya sejak ekonomi Indonesia terintegrasi ke dalam kapitalisme global. Guncangan sekecil apa pun pada lingkungan strategik, dipastikan berdampak signifikan terhadap kinerja perekonomian Indonesia. Fluktuasi harga komoditas di bursa berjangka internasional, suka atau tidak, pasti berpengaruh terbentuknya harga pada tingkat pertani di pedesaan Jatim.</p>
<p>Akibat kemarau panjang yang mendera, produksi gula India merosot drastis hingga berkurang sekitar 10,0 jutaton. Kalau biasanya negara anak benua ini mengekspor sekitar 3,0 juta ton, tahun ini terpaksa mengimpor gula dari pasar global. Stok gula dunia terguncang, apalagi ditambah dengan kembali menguatnya harga minyak bumi yang memaksa Brasil mengurangi produksi gula dan mengalihkannya sebagian untuk bioetanol. Maka untuk pertama kalinya dalam tiga dekade terakhir, harga gula dunia mencapai diatas harga 550 dollar AS per ton FOB (Harga di negara asal, belum termasuk biaya pengapalan dan premium).</p>
<p><!--more-->Keberadaan Indonesia sebagai produsen sekaligus importer gula menjadikan harga gula dunia menimbulkan dampak serius terhadap terbentuknya harga di pasar domestik. Sejaih ini, Indonesia baru bisa berswasembada gula konsumsi, sementara gula untuk bahan baku industri masih harus diimpor atau dibeli dari produsen lokal tetapi bahan baku raw sugar masih diimpor pula. Begitu harga dunia tidak lagi kompromi, impor tidak berjalan normal.</p>
<p><img class="alignleft size-thumbnail wp-image-2456" title="anomali hrg gula" src="http://hagemman.wordpress.com/files/2009/09/anomali-hrg-gula.jpg?w=150" alt="anomali hrg gula" width="150" height="126" />Jalan keluarnya yang dapat ditempuh bagi industri makanan/minuman untuk keluar dari perangkap situasi demikian adalah melakukan substitusi bahan baku dari gula rafinasi ke gula lokal. Rebutan gula lokal antara untuk konsumsi dan bahan baku industri menjadikan harga tak dapat dikendalikan. Imbauan pemerintah kepada BUMN agar harga gula pada level tender tidak lebih dari Rp 6.500 per kg berujuan supaya harga pada tingkat konsumen Rp 7.000 – Rp 7.500 pun bergeming.</p>
<p>Secara teoritik, harga tidak dapat didikte pemerintah. Namun, bukan berarti pemerintah tidak dapat melakukan intervensi. Intervensi pun boleh saja dilakukan untuk menyehatkan pasar. Namun untuk keperluan tersebut pemerintah harus menguasai stok. Cara efektif yang dapat ditempuh adalah membeli gula dari produsen dengan harga sesuai mekanisme pasar dan melepasnya ke konsumen dengan harga yang dikehendaki,</p>
<p>Cara semacam ini juga lazim ditempuh banyak negara industri maju penganut dan penganjur mazhab kapitalisme. Namun, apakah pemerintaj punya dana dan geregt melakukan intervensi. Naiknya harga gula di luar batas kewajaran memang berkah bagi petani dan diharapkan menimbulkan rasa percaya diri untuk terus meningkatkan daya saing komoditas usaha taninya. Kelihan muncul karena bangsa ini terbiasa tidak menghargai petani.</p>
<p>Harga komoditas agribisnis ditekan serendah mungkin meski untuk menjaga gawang ketahanan pangan sangat diperlukan insentif bagi pengelola usaha tani. Faktor lainnya, Indonesia terlanjur masuk perangkap impor. Investasi pabrik gula rafinasi secara besar-besaran tanpa upaya membangun kebun tebu adalah salah satu ealpaan tersebsar yang memberikan kontribusi terhadap melambungnya harga gula. Seruan petani tebu dan PG agar pemerintah segera mewajibkan semua industri gula rafinasi baik yang sudah eksis maupun yang berniat mengembangkan kapasitas dan investasi baru tidak juga digubris.</p>
<p>Lemahnya sinkronisasi antara industri gula rafinasi dan penggunanya sebagaimana tercermin dari masih berlanjutnya impor gula rafinasi sekali lagi memperkuat lemahnya integrasi dalam formulasi dan implementasi kebijakan pergulaan nasional. Baru sekarang setelah harga gula melambung dan menimbulkan anomali, semuanya terasa. Akankah kebijakan pergulaan dibiarkan setelah kasus ini ? Semuanya tergantung pada komitmen negara apakah mau didikte fluktuasi harga dunia yang tidak pernah jelas atau berpihak kepada petani dengan menata ulang semua kebijakan kontraproduktif terhadao pemberdayaannya ?</p>
<p>Sumber  :</p>
<p>Di Balik Anomali Harga Gula, Adig Suwandi &#124; Praktisi Agribisnis, Alumnus Universitas Brawijaya Malang<br />
Kompas Jawa Timur, 01.09.2009</p>
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<item>
<title><![CDATA[Substitusi Komponen Satria FU]]></title>
<link>http://ekojuwono.wordpress.com/2009/07/31/substitusi-komponen-satria-fu/</link>
<pubDate>Fri, 31 Jul 2009 00:48:55 +0000</pubDate>
<dc:creator>sekopati</dc:creator>
<guid>http://ekojuwono.wordpress.com/2009/07/31/substitusi-komponen-satria-fu/</guid>
<description><![CDATA[Anda pemilik Suzuki Satria FU150? Mesinnya mengeluarkan suara berisik; duk..duk..duk.., terus tarika]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>Anda pemilik Suzuki Satria FU150? Mesinnya mengeluarkan suara berisik; duk..duk..duk.., terus tarikannya pun melempem? Wah, mesti servis besar tuh. Apalagi jika umur pemakaian besutan sudah 3 tahun lebih, buru-buru deh diperiksa sebelum terlambat.</p>
<p>Sebenarnya suara itu dari mana sih? “Penyebabnya ada dua. Pertama kalau laju motor ogah ngibrit, bisa diakibatkan kampas kopling aus, tapi kalau kampasnya masih bagus dan suara kasar itu masih ada, tinggal kita periksa di pen pistonnya,” ungkap Kiki, mekanik sekaligus pemilik bengkel Rizki Motor (RM).<!--more--></p>
<p>Menurut mekanik humoris ini, penyebab yang disebut nomor dua itu sering sekali terjadi. Maklum pen piston itu sebagai jembatan antara piston dan setang seher yang bekerja naik turun saat mesin hidup. Kalo peranti itu bermasalah, akan menimbulkan bunyi berisik, lama-lama bisa bikin bengkok setang sehernya dan yang paling fatal bisa patah akibat dipaksa bekerja. Hii serem..!</p>
<p>Contoh kasus dialami Artur Akihari warga Cipulir, Kebayoran Lama, Jaksel pada besutannya. Berhubung hidupnya masih bergantung pada orang tua, tentu dana yang dimiliki sangat terbatas, “Apa ada solusi lain Bang? Dana ogut cekak, nih!” harapnya cemas. </p>
<p>“Tenang Bung. Bisa kita pakai pen piston punya Yamaha RX-King, selain harganya murah; Rp 20 ribu dari harga asli FU Rp 35 ribu, bahan dan kualitasnya juga sama,” terang Kiki yang berbadan besar. Jadi penasaran, nih?</p>
<p>Setelah diukur sigmat, tercatat kalau pen piston FU punya panjang 49 mm dengan ketebalan 2,5 mm. Sedang punya RX-King lebih pendek 0,2 mm dari FU yaitu 47 mm dan memiliki tebal 2 mm, hanya diameter lingkarnya saja yang sama-sama 16 mm . Tapi ukuran panjangnya kan beda?</p>
<p>“Asalkan diameter lingkar pennya sama, gak ada masalah kok dan kalau kita lihat dari perbedaan ukuran ketebalannya 0,5 mm antara kedua pen tadi, sudah pasti mengurangi beban kerja mesin, akselerasi pun mengalami perubahan sedikit ,” beber pria yang doyan pakai topi ini. </p>
<p>Apa semudah itu masangnya? “Betul, sama kayak waktu ngelepas, yang paling penting adalah patokan klep in dan klep outnya jangan sampai salah, sebab kalau salah, masing-masing klep bisa tabrakan,” wantinya sembari bilang untuk pengerjaan tentu bisa minta bantuan mekanik bengkel .</p>
<p>Sumber : otomotifnet.com</p>
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<item>
<title><![CDATA[Satu arah!]]></title>
<link>http://ibn07.wordpress.com/2009/06/20/satu-arah/</link>
<pubDate>Sat, 20 Jun 2009 15:15:17 +0000</pubDate>
<dc:creator>Mantaray</dc:creator>
<guid>http://ibn07.wordpress.com/2009/06/20/satu-arah/</guid>
<description><![CDATA[cinta, orang billang itu indah, penuh inspirasi dan membangkitan semangat. iyakah? apa jadinya jika ]]></description>
<content:encoded><![CDATA[cinta, orang billang itu indah, penuh inspirasi dan membangkitan semangat. iyakah? apa jadinya jika ]]></content:encoded>
</item>
<item>
<title><![CDATA[Solusi Persamaan Linier]]></title>
<link>http://rincikembang.wordpress.com/2009/04/23/31/</link>
<pubDate>Thu, 23 Apr 2009 08:31:46 +0000</pubDate>
<dc:creator>rincikembang</dc:creator>
<guid>http://rincikembang.wordpress.com/2009/04/23/31/</guid>
<description><![CDATA[Penyelesain Persamaan Linier]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p><a title="Penyelesaian Persamaan Linier" href="http://www.ziddu.com/download/4381306/Ke-2PENYELESAIANPERSAMAANLINIE.PPT.html" target="_blank">Penyelesain Persamaan Linier</a></p>
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</item>
<item>
<title><![CDATA[Kanada 1971 #4]]></title>
<link>http://olimpiadematematika.wordpress.com/2009/04/10/kanada-1971-4-2/</link>
<pubDate>Fri, 10 Apr 2009 02:39:30 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://olimpiadematematika.wordpress.com/2009/04/10/kanada-1971-4-2/</guid>
<description><![CDATA[4. Cari semua bilangan real sehingga dua polinomial dan memiliki minimal satu akar yang sama. Solusi]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>4. Cari semua bilangan real <img src='http://l.wordpress.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' /> sehingga dua polinomial <img src='http://l.wordpress.com/latex.php?latex=x%5E2%2Bax%2B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+ax+1' title='x^2+ax+1' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=x%5E2%2Bx%2Ba&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+x+a' title='x^2+x+a' class='latex' /> memiliki minimal satu akar yang sama.</p>
<p>Solusi:</p>
<p>Misalkan <img src='http://l.wordpress.com/latex.php?latex=x%5E2%2Bax%2B1%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+ax+1=0' title='x^2+ax+1=0' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=x%5E2%2Bx%2Ba&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+x+a' title='x^2+x+a' class='latex' /> di mana <img src='http://l.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' /> adalah suatu bilangan real. Kurangi yang kedua dari yang pertama, <img src='http://l.wordpress.com/latex.php?latex=%28a-1%29x%2B%281-a%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(a-1)x+(1-a)=0' title='(a-1)x+(1-a)=0' class='latex' />, sehingga <img src='http://l.wordpress.com/latex.php?latex=%28a-1%29x%3D%28a-1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(a-1)x=(a-1)' title='(a-1)x=(a-1)' class='latex' />. Jika <img src='http://l.wordpress.com/latex.php?latex=a%5Cne1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a\ne1' title='a\ne1' class='latex' />, maka <img src='http://l.wordpress.com/latex.php?latex=x%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=1' title='x=1' class='latex' />. Substitusikan ke persamaan pertama, <img src='http://l.wordpress.com/latex.php?latex=1%2Ba%2B1%3D0%2Ca%3D-2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1+a+1=0,a=-2' title='1+a+1=0,a=-2' class='latex' />. Jika <img src='http://l.wordpress.com/latex.php?latex=a%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a=1' title='a=1' class='latex' />, jelas bahwa kedua polinomial sama, sehingga akar-akarnya pasti sama. Jadi nilai <img src='http://l.wordpress.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' /> adalah -2,1.</p>
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<item>
<title><![CDATA[Persamaan fungsional]]></title>
<link>http://artofmathematics.wordpress.com/2008/06/26/persamaan-fungsional-2/</link>
<pubDate>Thu, 26 Jun 2008 11:04:14 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/06/26/persamaan-fungsional-2/</guid>
<description><![CDATA[[Makedonia 2007] Tentukan semua fungsi yang memenuhi untuk setiap . Solusi memberikan . memberikan s]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p style="text-align:left;">[Makedonia 2007] Tentukan semua fungsi <img src='http://l.wordpress.com/latex.php?latex=f%3A%5Cmathbb%7BR%7D%5Crightarrow%5Cmathbb%7BR%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f:\mathbb{R}\rightarrow\mathbb{R}' title='f:\mathbb{R}\rightarrow\mathbb{R}' class='latex' /> yang memenuhi</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=+++f+%28x%5E%7B3%7D%2By%5E%7B3%7D%29+%3D+x%5E%7B2%7Df+%28x%29%2Byf+%28y%5E%7B2%7D%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='   f (x^{3}+y^{3}) = x^{2}f (x)+yf (y^{2})' title='   f (x^{3}+y^{3}) = x^{2}f (x)+yf (y^{2})' class='latex' /></p>
<p>untuk setiap <img src='http://l.wordpress.com/latex.php?latex=x%2Cy%5Cin%5Cmathbb%7BR%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x,y\in\mathbb{R}' title='x,y\in\mathbb{R}' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
<img src='http://l.wordpress.com/latex.php?latex=x%3D0%2Cy%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=0,y=0' title='x=0,y=0' class='latex' /> memberikan <img src='http://l.wordpress.com/latex.php?latex=f%280%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(0)=0' title='f(0)=0' class='latex' />.</p>
<p><img src='http://l.wordpress.com/latex.php?latex=y%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y=0' title='y=0' class='latex' /> memberikan <img src='http://l.wordpress.com/latex.php?latex=f%28x%5E3%29%3Dx%5E2f%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x^3)=x^2f(x)' title='f(x^3)=x^2f(x)' class='latex' /> sedangkan <img src='http://l.wordpress.com/latex.php?latex=x%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=0' title='x=0' class='latex' /> memberikan <img src='http://l.wordpress.com/latex.php?latex=f%28y%5E3%29%3Dyf%28y%29%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(y^3)=yf(y)^2' title='f(y^3)=yf(y)^2' class='latex' />.</p>
<p>Jadi <img src='http://l.wordpress.com/latex.php?latex=f%28x%5E3%2By%5E3%29%3Df%28x%5E3%29%2Bf%28y%5E3%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x^3+y^3)=f(x^3)+f(y^3)' title='f(x^3+y^3)=f(x^3)+f(y^3)' class='latex' /> atau <img src='http://l.wordpress.com/latex.php?latex=f%28z%2Bw%29%3Df%28z%29%2Bf%28w%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(z+w)=f(z)+f(w)' title='f(z+w)=f(z)+f(w)' class='latex' />.</p>
<p>Karena <img src='http://l.wordpress.com/latex.php?latex=x%5E2f%28x%29%3Dxf%28x%5E2%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2f(x)=xf(x^2)' title='x^2f(x)=xf(x^2)' class='latex' />, maka <img src='http://l.wordpress.com/latex.php?latex=f%28x%5E2%29%3Dxf%28x%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x^2)=xf(x)' title='f(x^2)=xf(x)' class='latex' /> untuk <img src='http://l.wordpress.com/latex.php?latex=x%5Cne0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x\ne0' title='x\ne0' class='latex' />.</p>
<p>Untuk <img src='http://l.wordpress.com/latex.php?latex=x%3D%5Cne-1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=\ne-1' title='x=\ne-1' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=f%28%28x%2B1%29%5E2%29%3D%28x%2B1%29f%28x%2B1%29%3D%28x%2B1%29%28f%28x%29%2Bf%281%29%29%3Dxf%28x%29%2Bxf%281%29%2Bf%28x%29%2Bf%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f((x+1)^2)=(x+1)f(x+1)=(x+1)(f(x)+f(1))=xf(x)+xf(1)+f(x)+f(1)' title='f((x+1)^2)=(x+1)f(x+1)=(x+1)(f(x)+f(1))=xf(x)+xf(1)+f(x)+f(1)' class='latex' />.</p>
<p>Tetapi <img src='http://l.wordpress.com/latex.php?latex=f%28%28x%2B1%29%5E2%29%3Df%28x%5E2%2Bx%2Bx%2B1%29%3Df%28x%5E2%29%2Bf%28x%29%2Bf%28x%29%2Bf%281%29%3Dxf%28x%29%2B2f%28x%29%2Bf%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f((x+1)^2)=f(x^2+x+x+1)=f(x^2)+f(x)+f(x)+f(1)=xf(x)+2f(x)+f(1)' title='f((x+1)^2)=f(x^2+x+x+1)=f(x^2)+f(x)+f(x)+f(1)=xf(x)+2f(x)+f(1)' class='latex' />.</p>
<p>Kedua persamaan terakhir menyebabkan <img src='http://l.wordpress.com/latex.php?latex=f%28x%29%3Dxf%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x)=xf(1)' title='f(x)=xf(1)' class='latex' /> untuk <img src='http://l.wordpress.com/latex.php?latex=x%5Cne0%2C-1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x\ne0,-1' title='x\ne0,-1' class='latex' />. Perhatikan bahwa <img src='http://l.wordpress.com/latex.php?latex=f%280%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(0)=0' title='f(0)=0' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=f%28-1%29%3D-f%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(-1)=-f(1)' title='f(-1)=-f(1)' class='latex' />. Jadi <img src='http://l.wordpress.com/latex.php?latex=f%28x%29%3Dxf%281%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x)=xf(1)' title='f(x)=xf(1)' class='latex' /> untuk setiap bilangan real <img src='http://l.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' />. Jadi <img src='http://l.wordpress.com/latex.php?latex=f%28x%29%3Dax&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x)=ax' title='f(x)=ax' class='latex' /> untuk suatu konstanta <img src='http://l.wordpress.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' />.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Hasil kali akar-akar]]></title>
<link>http://artofmathematics.wordpress.com/2008/06/19/hasil-kali-akar-akar/</link>
<pubDate>Thu, 19 Jun 2008 16:20:41 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/06/19/hasil-kali-akar-akar/</guid>
<description><![CDATA[[AIME 1983] Tentukan hasil kali dari akar-akar real persamaan . Solusi Misalkan , sehingga . Kuadrat]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[AIME 1983] Tentukan hasil kali dari akar-akar real persamaan <img src='http://l.wordpress.com/latex.php?latex=x%5E2%2B18x%2B30%3D2%5Csqrt%7Bx%5E2%2B18x%2B45%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+18x+30=2\sqrt{x^2+18x+45}' title='x^2+18x+30=2\sqrt{x^2+18x+45}' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Misalkan <img src='http://l.wordpress.com/latex.php?latex=y%3Dx%5E2%2B18x%2B30&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y=x^2+18x+30' title='y=x^2+18x+30' class='latex' />, sehingga <img src='http://l.wordpress.com/latex.php?latex=y%3D2%5Csqrt%7By%2B15%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y=2\sqrt{y+15}' title='y=2\sqrt{y+15}' class='latex' />. Kuadratkan menjadi <img src='http://l.wordpress.com/latex.php?latex=y%5E2%3D4y%2B60&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y^2=4y+60' title='y^2=4y+60' class='latex' /> atau <img src='http://l.wordpress.com/latex.php?latex=y%5E2-4y-60%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y^2-4y-60=0' title='y^2-4y-60=0' class='latex' />, yang menyebabkan <img src='http://l.wordpress.com/latex.php?latex=%28y%2B6%29%28y-10%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(y+6)(y-10)=0' title='(y+6)(y-10)=0' class='latex' />. Jika <img src='http://l.wordpress.com/latex.php?latex=y%3D-6&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y=-6' title='y=-6' class='latex' />, maka <img src='http://l.wordpress.com/latex.php?latex=x%5E2%2B18x%2B36%3D-6&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+18x+36=-6' title='x^2+18x+36=-6' class='latex' />, yang tidak memiliki akar real. Jika <img src='http://l.wordpress.com/latex.php?latex=y%3D10&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y=10' title='y=10' class='latex' />, maka <img src='http://l.wordpress.com/latex.php?latex=x%5E2%2B18x%2B20%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+18x+20=0' title='x^2+18x+20=0' class='latex' />, yang memiliki hasil kali akar-akar <img src='http://l.wordpress.com/latex.php?latex=10&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='10' title='10' class='latex' />.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Persamaan fungsional]]></title>
<link>http://artofmathematics.wordpress.com/2008/06/08/persamaan-fungsional/</link>
<pubDate>Sun, 08 Jun 2008 16:28:16 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/06/08/persamaan-fungsional/</guid>
<description><![CDATA[[HMMT 2008] Fungsi memenuhi untuk bilangan real . Tentukan nilai dari . Solusi Substitusi , sehingga]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[HMMT 2008] Fungsi <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f' title='f' class='latex' /> memenuhi <img src='http://l.wordpress.com/latex.php?latex=f%28x%29+%2B+f%282x+%2B+y%29+%2B+5xy+%3D+f%283x+-+y%29+%2B+2x%5E2+%2B+1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(x) + f(2x + y) + 5xy = f(3x - y) + 2x^2 + 1' title='f(x) + f(2x + y) + 5xy = f(3x - y) + 2x^2 + 1' class='latex' /> untuk bilangan real <img src='http://l.wordpress.com/latex.php?latex=x%2Cy&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x,y' title='x,y' class='latex' />. Tentukan nilai dari <img src='http://l.wordpress.com/latex.php?latex=f%2810%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(10)' title='f(10)' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Substitusi <img src='http://l.wordpress.com/latex.php?latex=x%3D10%2Cy%3D5&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=10,y=5' title='x=10,y=5' class='latex' />, sehingga <img src='http://l.wordpress.com/latex.php?latex=f%2810%29%2Bf%2825%29%2B250%3Df%2825%29%2B200%2B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(10)+f(25)+250=f(25)+200+1' title='f(10)+f(25)+250=f(25)+200+1' class='latex' />, sehingga <img src='http://l.wordpress.com/latex.php?latex=f%2810%29%3D-49&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(10)=-49' title='f(10)=-49' class='latex' />.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Garis dividen segitiga]]></title>
<link>http://artofmathematics.wordpress.com/2008/04/20/garis-dividen-segitiga/</link>
<pubDate>Sun, 20 Apr 2008 05:40:23 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/04/20/garis-dividen-segitiga/</guid>
<description><![CDATA[[GMO - Olimpiade.org] Katakan sebuah garis dalam segitiga dividen apabila ditarik dari suatu sudut s]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[GMO - Olimpiade.org] Katakan sebuah garis dalam segitiga dividen apabila ditarik dari suatu sudut segitiga, dan membagi<br />
segitiga menjadi dua bagian dengan keliling sama.<br />
Buktikan ketiga dividen suatu segitiga selalu berpotongan di satu titik !</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Misalkan sisi-sisi segitiga itu adalah <img src='http://l.wordpress.com/latex.php?latex=a%2Bb&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a+b' title='a+b' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=c%2Bd&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c+d' title='c+d' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=e%2Bf&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e+f' title='e+f' class='latex' />, di mana <img src='http://l.wordpress.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='b' title='b' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=c&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c' title='c' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=d&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='d' title='d' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=e&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e' title='e' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f' title='f' class='latex' /> adalah segmen garis dari sisi segitiga yang terbagi oleh garis dividen. Perhatikan gambar berikut.</p>
<p style="text-align:center;"><img class="aligncenter size-medium wp-image-472" src="http://artofmathematics.wordpress.com/files/2008/04/dividensegitiga.gif" alt="" width="183" height="175" /></p>
<p>Maka didapat tiga persamaan, yaitu</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=e%2Bf%2Ba%3Dd%2Bc%2Bb&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e+f+a=d+c+b' title='e+f+a=d+c+b' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=f%2Ba%2Bb%3De%2Bd%2Bc&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f+a+b=e+d+c' title='f+a+b=e+d+c' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=a%2Bb%2Bc%3Dd%2Be%2Bf&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a+b+c=d+e+f' title='a+b+c=d+e+f' class='latex' />.</p>
<p>Dari persamaan satu dikurangi persamaan dua, maka <img src='http://l.wordpress.com/latex.php?latex=e-b%3Db-e&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e-b=b-e' title='e-b=b-e' class='latex' />, yang menyebabkan <img src='http://l.wordpress.com/latex.php?latex=e%3Db&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='e=b' title='e=b' class='latex' />. Dengan cara yang sama, dengan membandingkan persamaan-persamaan, didapat <img src='http://l.wordpress.com/latex.php?latex=a%3Dd&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a=d' title='a=d' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=c%3Df&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c=f' title='c=f' class='latex' />.</p>
<p>Menurut teorema Ceva, untuk membuktikan ketiga dividen berpotongan di satu titik, cukup dibuktikan bahwa <img src='http://l.wordpress.com/latex.php?latex=%5Cfrac%7Ba%7D%7Bb%7D%5Ccdot%5Cfrac%7Bc%7D%7Bd%7D%5Ccdot%5Cfrac%7Be%7D%7Bf%7D%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\frac{a}{b}\cdot\frac{c}{d}\cdot\frac{e}{f}=1' title='\frac{a}{b}\cdot\frac{c}{d}\cdot\frac{e}{f}=1' class='latex' />. Tetapi</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdfrac%7Ba%7D%7Bb%7D%5Ccdot%5Cdfrac%7Bc%7D%7Bd%7D%5Ccdot%5Cdfrac%7Be%7D%7Bf%7D%3D%5Cdfrac%7Ba%7D%7Bd%7D%5Ccdot%5Cdfrac%7Bc%7D%7Bf%7D%5Ccdot%5Cdfrac%7Be%7D%7Bb%7D%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{a}{b}\cdot\dfrac{c}{d}\cdot\dfrac{e}{f}=\dfrac{a}{d}\cdot\dfrac{c}{f}\cdot\dfrac{e}{b}=1' title='\dfrac{a}{b}\cdot\dfrac{c}{d}\cdot\dfrac{e}{f}=\dfrac{a}{d}\cdot\dfrac{c}{f}\cdot\dfrac{e}{b}=1' class='latex' />.</p>
<p>Maka ketiga dividen berpotongan di satu titik.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Solusi bulat persamaan]]></title>
<link>http://artofmathematics.wordpress.com/2008/04/13/solusi-bulat-fungsi/</link>
<pubDate>Sun, 13 Apr 2008 01:17:03 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/04/13/solusi-bulat-fungsi/</guid>
<description><![CDATA[[Mathematical Reflections 2006] Buktikan bahwa persamaan berikut tidak memiliki solusi dalam bilanga]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[Mathematical Reflections 2006] Buktikan bahwa persamaan berikut tidak memiliki solusi dalam bilangan bulat</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%28x-y%29%5E2%2B5%28x-y%29%2B25%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(x-y)^2+5(x-y)+25=0' title='(x-y)^2+5(x-y)+25=0' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Persamaan tersebut dapat diubah menjadi</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=x%5E2-%282y%2B5%29x%2B%28y%5E2-5y%2B25%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2-(2y+5)x+(y^2-5y+25)=0' title='x^2-(2y+5)x+(y^2-5y+25)=0' class='latex' />.</p>
<p>Diskriminan persamaan di atas adalah</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%282y%2B5%29%5E2-4%28y%5E2-5y%2B25%29%3D40y-75&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(2y+5)^2-4(y^2-5y+25)=40y-75' title='(2y+5)^2-4(y^2-5y+25)=40y-75' class='latex' />,</p>
<p>yang harus berupa bilangan kuadrat, agar nilai <img src='http://l.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' /> bulat. Maka <img src='http://l.wordpress.com/latex.php?latex=y&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y' title='y' class='latex' /> habis dibagi 5. Pada persamaan di atas, semua koefisien kecuali <img src='http://l.wordpress.com/latex.php?latex=x%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2' title='x^2' class='latex' /> habis dibagi 5, sehingga <img src='http://l.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' /> harus habis dibagi 5 juga. Maka misalkan <img src='http://l.wordpress.com/latex.php?latex=x%3D5x%27&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=5x&#039;' title='x=5x&#039;' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=y%3D5y%27&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y=5y&#039;' title='y=5y&#039;' class='latex' />. Substitusikan ini ke persamaan awal dan sederhanakan sehingga</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=x%27%5E2-%282y%27%2B1%29x%27%2B%28y%27%5E2-y%27%2B1%29%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x&#039;^2-(2y&#039;+1)x&#039;+(y&#039;^2-y&#039;+1)=0' title='x&#039;^2-(2y&#039;+1)x&#039;+(y&#039;^2-y&#039;+1)=0' class='latex' />.</p>
<p>Diskriminannya adalah</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%282y%27%2B1%29%5E2-4%28y%27%5E2-y%27%2B1%29%3D8y%27-3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(2y&#039;+1)^2-4(y&#039;^2-y&#039;+1)=8y&#039;-3' title='(2y&#039;+1)^2-4(y&#039;^2-y&#039;+1)=8y&#039;-3' class='latex' />,</p>
<p>yang tidak mungkin bilangan kuadrat. Jadi tidak ada solusi bilangan bulat.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Habis dibagi 30]]></title>
<link>http://artofmathematics.wordpress.com/2008/04/13/habis-dibagi-30/</link>
<pubDate>Sat, 12 Apr 2008 23:38:31 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/04/13/habis-dibagi-30/</guid>
<description><![CDATA[[Orisinil] Jika , , , , adalah bilangan-bilangan bulat yang jumlahnya 0, buktikan bahwa habis dibagi]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[Orisinil] Jika <img src='http://l.wordpress.com/latex.php?latex=a_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_1' title='a_1' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=a_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_2' title='a_2' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=a_3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_3' title='a_3' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cldots&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ldots' title='\ldots' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=a_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_n' title='a_n' class='latex' /> adalah bilangan-bilangan bulat yang jumlahnya 0, buktikan bahwa <img src='http://l.wordpress.com/latex.php?latex=a_1%5E5%2Ba_2%5E5%2Ba_3%5E5%2B%5Cldots%2Ba_n%5E5&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_1^5+a_2^5+a_3^5+\ldots+a_n^5' title='a_1^5+a_2^5+a_3^5+\ldots+a_n^5' class='latex' /> habis dibagi 30.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Perhatikan bahwa <img src='http://l.wordpress.com/latex.php?latex=a%5E5-a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^5-a' title='a^5-a' class='latex' /> habis dibagi 5, menurut teorema Fermat. Tetapi <img src='http://l.wordpress.com/latex.php?latex=a%5E5-a%3D%28a%5E2%2B1%29%28a%2B1%29a%28a-1%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^5-a=(a^2+1)(a+1)a(a-1)' title='a^5-a=(a^2+1)(a+1)a(a-1)' class='latex' />, sehingga memiliki faktor tiga bilangan berurutan, maka habis dibagi 6. Jadi <img src='http://l.wordpress.com/latex.php?latex=a%5E5-a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^5-a' title='a^5-a' class='latex' /> habis dibagi 30, atau <img src='http://l.wordpress.com/latex.php?latex=a%5E5%5Cequiv+a%5Cpmod%7B30%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^5\equiv a\pmod{30}' title='a^5\equiv a\pmod{30}' class='latex' />. Substitusikan untuk <img src='http://l.wordpress.com/latex.php?latex=a_1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_1' title='a_1' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=a_2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_2' title='a_2' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cldots&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\ldots' title='\ldots' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=a_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_n' title='a_n' class='latex' /> dan jumlahkan sehingga</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=a_1%5E5%2Ba_2%5E5%2B%5Cldots%2Ba_n%5E5%5Cequiv+a_1%2Ba_2%2B%5Cldots%2Ba_n%5Cpmod%7B30%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_1^5+a_2^5+\ldots+a_n^5\equiv a_1+a_2+\ldots+a_n\pmod{30}' title='a_1^5+a_2^5+\ldots+a_n^5\equiv a_1+a_2+\ldots+a_n\pmod{30}' class='latex' />.</p>
<p>Tetapi <img src='http://l.wordpress.com/latex.php?latex=a_1%2Ba_2%2B%5Cldots%2Ba_n%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_1+a_2+\ldots+a_n=0' title='a_1+a_2+\ldots+a_n=0' class='latex' />, sehingga <img src='http://l.wordpress.com/latex.php?latex=a_1%5E5%2Ba_2%5E5%2B%5Cldots%2Ba_n%5E5%5Cequiv0%5Cpmod%7B30%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_1^5+a_2^5+\ldots+a_n^5\equiv0\pmod{30}' title='a_1^5+a_2^5+\ldots+a_n^5\equiv0\pmod{30}' class='latex' />, dan terbukti.</p>
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</item>
<item>
<title><![CDATA[Determinan matriks]]></title>
<link>http://artofmathematics.wordpress.com/2008/04/05/determinan-matriks/</link>
<pubDate>Sat, 05 Apr 2008 08:08:46 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/04/05/determinan-matriks/</guid>
<description><![CDATA[[Putnam 2004] Didefinisikan barisan dengan , dan untuk setiap . Buktikan bahwa adalah bilangan bulat]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[Putnam 2004] Didefinisikan barisan <img src='http://l.wordpress.com/latex.php?latex=%5C%7Bu_n%5C%7D%5E%7B%5Cinfty%7D_%7Bn%3D0%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\{u_n\}^{\infty}_{n=0}' title='\{u_n\}^{\infty}_{n=0}' class='latex' /> dengan <img src='http://l.wordpress.com/latex.php?latex=u_0%3Du_1%3Du_2%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_0=u_1=u_2=0' title='u_0=u_1=u_2=0' class='latex' />, dan</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=%5Cdet%5Cleft%28%5Cbegin%7Barray%7D%7Bccc%7Du_n+%26%2338%3B+u_%7Bn%2B1%7D+%5C%5C+u_%7Bn%2B2%7D+%26%2338%3B+u_%7Bn%2B3%7D+%5Cend%7Barray%7D+%5Cright%29%3Dn%21&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\det\left(\begin{array}{ccc}u_n &amp; u_{n+1} \\ u_{n+2} &amp; u_{n+3} \end{array} \right)=n!' title='\det\left(\begin{array}{ccc}u_n &amp; u_{n+1} \\ u_{n+2} &amp; u_{n+3} \end{array} \right)=n!' class='latex' /></p>
<p>untuk setiap <img src='http://l.wordpress.com/latex.php?latex=n%5Cge0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\ge0' title='n\ge0' class='latex' />. Buktikan bahwa <img src='http://l.wordpress.com/latex.php?latex=u_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_n' title='u_n' class='latex' /> adalah bilangan bulat untuk setiap <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' />. (ditetapkan <img src='http://l.wordpress.com/latex.php?latex=0%21%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0!=1' title='0!=1' class='latex' />.)</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Saya definisikan <img src='http://l.wordpress.com/latex.php?latex=v_0%3Dv_1%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_0=v_1=1' title='v_0=v_1=1' class='latex' />, dan <img src='http://l.wordpress.com/latex.php?latex=v_n%3D%28n-1%29v_%7Bn-2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_n=(n-1)v_{n-2}' title='v_n=(n-1)v_{n-2}' class='latex' /> untuk setiap <img src='http://l.wordpress.com/latex.php?latex=n%5Cge2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\ge2' title='n\ge2' class='latex' />. Maka <img src='http://l.wordpress.com/latex.php?latex=v_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_n' title='v_n' class='latex' /> adalah bilangan bulat.</p>
<p><strong>Lemma 1.</strong> <img src='http://l.wordpress.com/latex.php?latex=v_nv_%7Bn%2B1%7D%3Dn%21&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_nv_{n+1}=n!' title='v_nv_{n+1}=n!' class='latex' /> untuk setiap <img src='http://l.wordpress.com/latex.php?latex=n%5Cge0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\ge0' title='n\ge0' class='latex' />.</p>
<p><em>Bukti</em><br />
Untuk <img src='http://l.wordpress.com/latex.php?latex=n%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n=0' title='n=0' class='latex' />, maka <img src='http://l.wordpress.com/latex.php?latex=v_nv_%7Bn%2B1%7D%3D1%3Dn%21&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_nv_{n+1}=1=n!' title='v_nv_{n+1}=1=n!' class='latex' />, seperti yang didefinisikan. Jika <img src='http://l.wordpress.com/latex.php?latex=n%5Cge1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\ge1' title='n\ge1' class='latex' />, asumsikan <img src='http://l.wordpress.com/latex.php?latex=v_%7Bn-1%7Dv_n%3D%28n-1%29%21&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_{n-1}v_n=(n-1)!' title='v_{n-1}v_n=(n-1)!' class='latex' />. Menurut definisi, <img src='http://l.wordpress.com/latex.php?latex=v_%7Bn%2B1%7D%3Dnv_%7Bn-1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_{n+1}=nv_{n-1}' title='v_{n+1}=nv_{n-1}' class='latex' />. Maka kedua ruas dikalikan dengan <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' />, sehingga <img src='http://l.wordpress.com/latex.php?latex=v_%7Bn%2B1%7Dv_n%3Dn%21&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_{n+1}v_n=n!' title='v_{n+1}v_n=n!' class='latex' />. Maka induksi selesai dan lemma terbukti.</p>
<p><strong>Lemma 2.</strong><img src='http://l.wordpress.com/latex.php?latex=v_n%3Du_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_n=u_n' title='v_n=u_n' class='latex' /> untuk setiap <img src='http://l.wordpress.com/latex.php?latex=n%5Cge0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\ge0' title='n\ge0' class='latex' />.</p>
<p><em>Bukti</em><br />
Untuk <img src='http://l.wordpress.com/latex.php?latex=n%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n=0' title='n=0' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=n%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n=1' title='n=1' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=n%3D2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n=2' title='n=2' class='latex' />, maka <img src='http://l.wordpress.com/latex.php?latex=u_n%3Dv_n%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_n=v_n=1' title='u_n=v_n=1' class='latex' />, seperti yang didefinisikan. Jika <img src='http://l.wordpress.com/latex.php?latex=n%5Cge1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\ge1' title='n\ge1' class='latex' />, asumsikan <img src='http://l.wordpress.com/latex.php?latex=u_%7Bn-3%7D%3Dv_%7Bn-3%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_{n-3}=v_{n-3}' title='u_{n-3}=v_{n-3}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=u_%7Bn-2%7D%3Dv_%7Bn-2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_{n-2}=v_{n-2}' title='u_{n-2}=v_{n-2}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=u_%7Bn-1%7D%3Dv_%7Bn-1%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_{n-1}=v_{n-1}' title='u_{n-1}=v_{n-1}' class='latex' />. Dari definisi determinan pada soal, maka <img src='http://l.wordpress.com/latex.php?latex=%28n-3%29%21%3Du_%7Bn-3%7Du_n-u_%7Bn-1%7Du_%7Bn-2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(n-3)!=u_{n-3}u_n-u_{n-1}u_{n-2}' title='(n-3)!=u_{n-3}u_n-u_{n-1}u_{n-2}' class='latex' />, maka</p>
<p style="text-align:center;"><img src='http://l.wordpress.com/latex.php?latex=u_n%3D%5Cdfrac%7B%28n-3%29%21%2Bu_%7Bn-1%7Du_%7Bn-2%7D%7D%7Bu_%7Bn-3%7D%7D%3D+%5Cdfrac%7B%28n-3%29%21%2B%28n-2%29v_%7Bn-3%7Dv_%7Bn-2%7D%7D%7Bv_%7Bn-3%7D%7D%3D%5Cdfrac%7Bv_%7Bn-3%7Dv_%7Bn-2%7D%2B%28n-2%29v_%7Bn-3%7Dv_%7Bn-2%7D%7D%7Bv_%7Bn-3%7D%7D%3D%28n-1%29v_%7Bn-2%7D%3Dv_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_n=\dfrac{(n-3)!+u_{n-1}u_{n-2}}{u_{n-3}}= \dfrac{(n-3)!+(n-2)v_{n-3}v_{n-2}}{v_{n-3}}=\dfrac{v_{n-3}v_{n-2}+(n-2)v_{n-3}v_{n-2}}{v_{n-3}}=(n-1)v_{n-2}=v_n' title='u_n=\dfrac{(n-3)!+u_{n-1}u_{n-2}}{u_{n-3}}= \dfrac{(n-3)!+(n-2)v_{n-3}v_{n-2}}{v_{n-3}}=\dfrac{v_{n-3}v_{n-2}+(n-2)v_{n-3}v_{n-2}}{v_{n-3}}=(n-1)v_{n-2}=v_n' class='latex' />.</p>
<p>Jadi <img src='http://l.wordpress.com/latex.php?latex=u_n%3Dv_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_n=v_n' title='u_n=v_n' class='latex' /> untuk setiap <img src='http://l.wordpress.com/latex.php?latex=n%5Cge0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\ge0' title='n\ge0' class='latex' />. Tetapi, karena <img src='http://l.wordpress.com/latex.php?latex=v_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='v_n' title='v_n' class='latex' /> selalu bilangan asli untuk <img src='http://l.wordpress.com/latex.php?latex=n%5Cge0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\ge0' title='n\ge0' class='latex' />, maka <img src='http://l.wordpress.com/latex.php?latex=u_n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='u_n' title='u_n' class='latex' /> juga selalu bilangan asli untuk setiap <img src='http://l.wordpress.com/latex.php?latex=n%5Cge0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\ge0' title='n\ge0' class='latex' />.</p>
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<item>
<title><![CDATA[Relasi Fungsional]]></title>
<link>http://artofmathematics.wordpress.com/2008/02/24/relasi-fungsional/</link>
<pubDate>Sun, 24 Feb 2008 10:37:25 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/02/24/relasi-fungsional/</guid>
<description><![CDATA[[Kosta Rika 2006] Diketahui adalah fungsi yang memenuhi . Tentukan . Solusi Substitusikan , maka . S]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[Kosta Rika 2006] Diketahui <img src='http://l.wordpress.com/latex.php?latex=f&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f' title='f' class='latex' /> adalah fungsi yang memenuhi</p>
<p align="center"> <img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+f%28x%29%2B2f%5Cleft%28%5Cdfrac%7Bx%2B%5Cdfrac%7B2001%7D%7B2%7D%7D%7Bx-1%7D%5Cright%29+%3D+4014-x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle f(x)+2f\left(\dfrac{x+\dfrac{2001}{2}}{x-1}\right) = 4014-x' title='\displaystyle f(x)+2f\left(\dfrac{x+\dfrac{2001}{2}}{x-1}\right) = 4014-x' class='latex' />.</p>
<p align="left">Tentukan <img src='http://l.wordpress.com/latex.php?latex=f%282004%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(2004)' title='f(2004)' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Substitusikan <img src='http://l.wordpress.com/latex.php?latex=x%3D2004&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=2004' title='x=2004' class='latex' />, maka</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+f%282004%29%2B2f%5Cleft%28%5Cdfrac32%5Cright%29%3D2010&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle f(2004)+2f\left(\dfrac32\right)=2010' title='\displaystyle f(2004)+2f\left(\dfrac32\right)=2010' class='latex' />.</p>
<p>Substitusikan <img src='http://l.wordpress.com/latex.php?latex=x%3D%5Cdfrac32&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x=\dfrac32' title='x=\dfrac32' class='latex' />, maka</p>
<p align="center"> <img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+f%5Cleft%28%5Cdfrac32%5Cright%29%2B2f%282004%29%3D4012%2C5&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle f\left(\dfrac32\right)+2f(2004)=4012,5' title='\displaystyle f\left(\dfrac32\right)+2f(2004)=4012,5' class='latex' />.</p>
<p>Jumlahkan dan bagi tiga</p>
<p align="center"> <img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle+f%282004%29%2Bf%5Cleft%28%5Cdfrac32%5Cright%29%3D2007%2C5&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle f(2004)+f\left(\dfrac32\right)=2007,5' title='\displaystyle f(2004)+f\left(\dfrac32\right)=2007,5' class='latex' />.</p>
<p>Kurangi persamaan terakhir dari persamaan kedua</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=f%282004%29%3D2005&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='f(2004)=2005' title='f(2004)=2005' class='latex' />.</p>
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<item>
<title><![CDATA[Akar polinomial kubik]]></title>
<link>http://artofmathematics.wordpress.com/2008/02/23/akar-polinomial-kubik/</link>
<pubDate>Sat, 23 Feb 2008 13:36:33 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/02/23/akar-polinomial-kubik/</guid>
<description><![CDATA[[wu :: forums] Tentukan polinomial berderajat tiga dengan akar bukan nol , , , sehingga adalah permu]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[wu :: forums] Tentukan polinomial berderajat tiga dengan akar bukan nol <img src='http://l.wordpress.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='b' title='b' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=c&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c' title='c' class='latex' />, sehingga <img src='http://l.wordpress.com/latex.php?latex=%28a%2Cb%2Cc%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(a,b,c)' title='(a,b,c)' class='latex' /> adalah permutasi dari <img src='http://l.wordpress.com/latex.php?latex=%28ab%2Bc%2Cbc%2Ba%2Cca%2Bb%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(ab+c,bc+a,ca+b)' title='(ab+c,bc+a,ca+b)' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Misalkan polinomial itu <img src='http://l.wordpress.com/latex.php?latex=%28x-a%29%28x-b%29%28x-c%29%3Dx%5E3-sx%5E2%2Btx-p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(x-a)(x-b)(x-c)=x^3-sx^2+tx-p' title='(x-a)(x-b)(x-c)=x^3-sx^2+tx-p' class='latex' />, sehingga <img src='http://l.wordpress.com/latex.php?latex=s%3Da%2Bb%2Bc&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s=a+b+c' title='s=a+b+c' class='latex' />,  <img src='http://l.wordpress.com/latex.php?latex=t%3Dab%2Bbc%2Bca&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t=ab+bc+ca' title='t=ab+bc+ca' class='latex' />, dan <img src='http://l.wordpress.com/latex.php?latex=p%3Dabc&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p=abc' title='p=abc' class='latex' />.</p>
<p>Karena <img src='http://l.wordpress.com/latex.php?latex=a%2Bb%2Bc%3D%28ab%2Bc%29%2B%28bc%2Ba%29%2B%28ca%2Bb%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a+b+c=(ab+c)+(bc+a)+(ca+b)' title='a+b+c=(ab+c)+(bc+a)+(ca+b)' class='latex' />, maka <img src='http://l.wordpress.com/latex.php?latex=s%3Dt%2Bs&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s=t+s' title='s=t+s' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=t%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t=0' title='t=0' class='latex' />.</p>
<p>Kemudian perhatikan bahwa</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=ab%2Bbc%2Bca%3D%28ab%2Bc%29%28bc%2Ba%29%2B%28bc%2Ba%29%28ca%2Bb%29%2B%28ca%2Bb%29%28ab%2Bc%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='ab+bc+ca=(ab+c)(bc+a)+(bc+a)(ca+b)+(ca+b)(ab+c)' title='ab+bc+ca=(ab+c)(bc+a)+(bc+a)(ca+b)+(ca+b)(ab+c)' class='latex' />.</p>
<p>Maka</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=t%3D%28ab%5E2c%2Bbc%5E2%2Ba%5E2b%2Bca%29%2B%28abc%5E2%2Ba%5E2c%2Bb%5E2c%2Ba%5E2c%2Bab%29%2B%28a%5E2bc%2Bab%5E2%2Bac%5E2%2Bbc%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t=(ab^2c+bc^2+a^2b+ca)+(abc^2+a^2c+b^2c+a^2c+ab)+(a^2bc+ab^2+ac^2+bc)' title='t=(ab^2c+bc^2+a^2b+ca)+(abc^2+a^2c+b^2c+a^2c+ab)+(a^2bc+ab^2+ac^2+bc)' class='latex' />.</p>
<p> Dengan beberapa substitusi dan penyederhanaan, didapat</p>
<p align="center"> <img src='http://l.wordpress.com/latex.php?latex=t%3Dps%2Bst-3p%2Bt&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t=ps+st-3p+t' title='t=ps+st-3p+t' class='latex' />.</p>
<p> Karena <img src='http://l.wordpress.com/latex.php?latex=t%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='t=0' title='t=0' class='latex' />, maka</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=0%3Dps-3p%3Dp%28s-3%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0=ps-3p=p(s-3)' title='0=ps-3p=p(s-3)' class='latex' />.</p>
<p>Tetapi <img src='http://l.wordpress.com/latex.php?latex=p%3Dabc%5Cne0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p=abc\ne0' title='p=abc\ne0' class='latex' />, sehingga <img src='http://l.wordpress.com/latex.php?latex=s%3D3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='s=3' title='s=3' class='latex' />.</p>
<p><img src='http://l.wordpress.com/latex.php?latex=p%3Dabc%3D%28ab%2Bc%29%28bc%2Ba%29%28ca%2Bb%29%3Dp%5E2%2Bp%28s%5E2-2t%29%2B%28t%5E2-2sp%29%2Bp&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p=abc=(ab+c)(bc+a)(ca+b)=p^2+p(s^2-2t)+(t^2-2sp)+p' title='p=abc=(ab+c)(bc+a)(ca+b)=p^2+p(s^2-2t)+(t^2-2sp)+p' class='latex' />. Selesaikan sehingga <img src='http://l.wordpress.com/latex.php?latex=p%3D-3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p=-3' title='p=-3' class='latex' />.</p>
<p>Maka polinomial itu adalah <img src='http://l.wordpress.com/latex.php?latex=x%5E3%2B3x%5E2-3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^3+3x^2-3' title='x^3+3x^2-3' class='latex' />.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Diagonal tegak lurus]]></title>
<link>http://artofmathematics.wordpress.com/2008/02/09/diagonal-tegak-lurus/</link>
<pubDate>Fri, 08 Feb 2008 23:42:15 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/02/09/diagonal-tegak-lurus/</guid>
<description><![CDATA[[Belanda 1998] adalah segi empat konveks sehingga . (a) Buktikan . (b) adalah segi empat konveks seh]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[Belanda 1998] <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BABCD%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{ABCD}' title='\text{ABCD}' class='latex' /> adalah segi empat konveks sehingga <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BAC%7D%5Cperp%5Ctext%7BBD%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{AC}\perp\text{BD}' title='\text{AC}\perp\text{BD}' class='latex' />.</p>
<p>(a) Buktikan <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BAB%7D%5E2%2B%5Ctext%7BCD%7D%5E2%3D%5Ctext%7BBC%7D%5E2%2B%5Ctext%7BAD%7D%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{AB}^2+\text{CD}^2=\text{BC}^2+\text{AD}^2' title='\text{AB}^2+\text{CD}^2=\text{BC}^2+\text{AD}^2' class='latex' />.</p>
<p>(b) <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BPQRS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{PQRS}' title='\text{PQRS}' class='latex' /> adalah segi empat konveks sehingga <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BPQ%7D%3D%5Ctext%7BAB%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{PQ}=\text{AB}' title='\text{PQ}=\text{AB}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BQR%7D%3D%5Ctext%7BBC%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{QR}=\text{BC}' title='\text{QR}=\text{BC}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BRS%7D%3D%5Ctext%7BCD%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{RS}=\text{CD}' title='\text{RS}=\text{CD}' class='latex' />, dan <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BSP%7D%3D%5Ctext%7BDA%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{SP}=\text{DA}' title='\text{SP}=\text{DA}' class='latex' />. Buktikan <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BPR%7D%5Cperp%5Ctext%7BQS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{PR}\perp\text{QS}' title='\text{PR}\perp\text{QS}' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
(a) Misalkan <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BAC%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{AC}' title='\text{AC}' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BBD%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{BD}' title='\text{BD}' class='latex' /> berpotongan di <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BO%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{O}' title='\text{O}' class='latex' />.</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BAB%7D%5E2%2B%5Ctext%7BCD%7D%5E2%3D%5Ctext%7BOA%7D%5E2%2B%5Ctext%7BOB%7D%5E2%2B%5Ctext%7BOC%7D%5E2%2B%5Ctext%7BOD%7D%5E2%3D%5Ctext%7BBC%7D%5E2%2B%5Ctext%7BAD%7D%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{AB}^2+\text{CD}^2=\text{OA}^2+\text{OB}^2+\text{OC}^2+\text{OD}^2=\text{BC}^2+\text{AD}^2' title='\text{AB}^2+\text{CD}^2=\text{OA}^2+\text{OB}^2+\text{OC}^2+\text{OD}^2=\text{BC}^2+\text{AD}^2' class='latex' />.</p>
<p>(b) Substitusikan nilai <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BPQ%7D%3D%5Ctext%7BAB%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{PQ}=\text{AB}' title='\text{PQ}=\text{AB}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BQR%7D%3D%5Ctext%7BBC%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{QR}=\text{BC}' title='\text{QR}=\text{BC}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BRS%7D%3D%5Ctext%7BCD%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{RS}=\text{CD}' title='\text{RS}=\text{CD}' class='latex' />, dan <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BSP%7D%3D%5Ctext%7BDA%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{SP}=\text{DA}' title='\text{SP}=\text{DA}' class='latex' /> ke persamaan yang telah dibuktikan (a). Maka <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BPQ%7D%5E2%2B%5Ctext%7BRS%7D%5E2%3D%5Ctext%7BQR%7D%5E2%2B%5Ctext%7BSP%7D%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{PQ}^2+\text{RS}^2=\text{QR}^2+\text{SP}^2' title='\text{PQ}^2+\text{RS}^2=\text{QR}^2+\text{SP}^2' class='latex' />.</p>
<p>Misalkan <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BPR%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{PR}' title='\text{PR}' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BQS%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{QS}' title='\text{QS}' class='latex' /> berpotongan di <img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BM%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{M}' title='\text{M}' class='latex' />. Tanpa mengurangi keumuman, dapat diasumsikan <img src='http://l.wordpress.com/latex.php?latex=%5Cangle+%5Ctext%7BPMQ%7D%3D%5Cangle+%5Ctext%7BRMQ%7D%5Cge90%5E%5Ccirc&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\angle \text{PMQ}=\angle \text{RMQ}\ge90^\circ' title='\angle \text{PMQ}=\angle \text{RMQ}\ge90^\circ' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=%5Cangle+%5Ctext%7BRMQ%7D%3D%5Cangle+%5Ctext%7BPMS%7D%5Cle90%5E%5Ccirc&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\angle \text{RMQ}=\angle \text{PMS}\le90^\circ' title='\angle \text{RMQ}=\angle \text{PMS}\le90^\circ' class='latex' />.</p>
<p>Maka</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BPQ%7D%5E2%5Cge+%5Ctext%7BPM%7D%5E2%2B%5Ctext%7BMQ%7D%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{PQ}^2\ge \text{PM}^2+\text{MQ}^2' title='\text{PQ}^2\ge \text{PM}^2+\text{MQ}^2' class='latex' />,</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BRS%7D%5E2%5Cge%5Ctext%7BMS%7D%5E2%2B%5Ctext%7BMR%7D%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{RS}^2\ge\text{MS}^2+\text{MR}^2' title='\text{RS}^2\ge\text{MS}^2+\text{MR}^2' class='latex' />,</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BQR%7D%5E2%5Cle+%5Ctext%7BMQ%7D%5E2%2B%5Ctext%7BMR%7D%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{QR}^2\le \text{MQ}^2+\text{MR}^2' title='\text{QR}^2\le \text{MQ}^2+\text{MR}^2' class='latex' />,</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BPS%7D%5E2%5Cle+%5Ctext%7BPM%7D%5E2%2B%5Ctext%7BMS%7D%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{PS}^2\le \text{PM}^2+\text{MS}^2' title='\text{PS}^2\le \text{PM}^2+\text{MS}^2' class='latex' />.</p>
<p align="left">Jadi</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Ctext%7BPQ%7D%5E2%2B%5Ctext%7BRS%7D%5E2%5Cge+%5Ctext%7BPM%7D%5E2%2B%5Ctext%7BMQ%7D%5E2%2B%5Ctext%7BMS%7D%5E2%2B%5Ctext%7BMR%7D%5E2%5Cge+%5Ctext%7BQR%7D%5E2%2B%5Ctext%7BPS%7D%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\text{PQ}^2+\text{RS}^2\ge \text{PM}^2+\text{MQ}^2+\text{MS}^2+\text{MR}^2\ge \text{QR}^2+\text{PS}^2' title='\text{PQ}^2+\text{RS}^2\ge \text{PM}^2+\text{MQ}^2+\text{MS}^2+\text{MR}^2\ge \text{QR}^2+\text{PS}^2' class='latex' />.</p>
<p>Tetapi persamaan memenuhi, sehingga</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cangle+%5Ctext%7BPMQ%7D%3D%5Cangle+%5Ctext%7BQMR%7D%3D%5Cangle%5Ctext%7BSMR%7D%3D%5Cangle+%5Ctext%7BSMP%7D%3D90%5E%5Ccirc&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\angle \text{PMQ}=\angle \text{QMR}=\angle\text{SMR}=\angle \text{SMP}=90^\circ' title='\angle \text{PMQ}=\angle \text{QMR}=\angle\text{SMR}=\angle \text{SMP}=90^\circ' class='latex' />.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Pertidaksamaan dua bilangan]]></title>
<link>http://artofmathematics.wordpress.com/2008/02/07/pertidaksamaan-dua-bilangan/</link>
<pubDate>Thu, 07 Feb 2008 00:38:46 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/02/07/pertidaksamaan-dua-bilangan/</guid>
<description><![CDATA[[Harold Shapiro] Jika dan adalah bilangan real positif, dan , buktikan . Solusi Misalkan . Karena da]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[Harold Shapiro] Jika <img src='http://l.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=y&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y' title='y' class='latex' /> adalah bilangan real positif, dan <img src='http://l.wordpress.com/latex.php?latex=0%26%2360%3Bp%26%2360%3B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='0&lt;p&lt;1' title='0&lt;p&lt;1' class='latex' />, buktikan <img src='http://l.wordpress.com/latex.php?latex=%28x%2By%29%5Ep%26%2360%3Bx%5Ep%2By%5Ep&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(x+y)^p&lt;x^p+y^p' title='(x+y)^p&lt;x^p+y^p' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Misalkan <img src='http://l.wordpress.com/latex.php?latex=x%2By%3Da&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x+y=a' title='x+y=a' class='latex' />. Karena <img src='http://l.wordpress.com/latex.php?latex=x%2Fa%26%2360%3B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x/a&lt;1' title='x/a&lt;1' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=p%26%2360%3B1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p&lt;1' title='p&lt;1' class='latex' />, maka</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdfrac%7Bx%7D%7Ba%7D%26%2360%3B%5Cdisplaystyle%5Cleft%28%5Cdfrac%7Bx%7D%7Ba%7D%5Cright%29%5Ep&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{x}{a}&lt;\displaystyle\left(\dfrac{x}{a}\right)^p' title='\dfrac{x}{a}&lt;\displaystyle\left(\dfrac{x}{a}\right)^p' class='latex' />.</p>
<p>Dengan logika yang sama, maka didapat</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdfrac%7By%7D%7Ba%7D%26%2360%3B%5Cdisplaystyle%5Cleft%28%5Cdfrac%7By%7D%7Ba%7D%5Cright%29%5Ep&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\dfrac{y}{a}&lt;\displaystyle\left(\dfrac{y}{a}\right)^p' title='\dfrac{y}{a}&lt;\displaystyle\left(\dfrac{y}{a}\right)^p' class='latex' />.</p>
<p>Maka</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=1%3D%5Cdfrac%7Bx%7D%7Ba%7D%2B%5Cdfrac%7By%7D%7Ba%7D%26%2360%3B%5Cdisplaystyle%5Cleft%28%5Cdfrac%7Bx%7D%7Ba%7D%5Cright%29%5Ep%2B%5Cdisplaystyle%5Cleft%28%5Cdfrac%7By%7D%7Ba%7D%5Cright%29%5Ep&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1=\dfrac{x}{a}+\dfrac{y}{a}&lt;\displaystyle\left(\dfrac{x}{a}\right)^p+\displaystyle\left(\dfrac{y}{a}\right)^p' title='1=\dfrac{x}{a}+\dfrac{y}{a}&lt;\displaystyle\left(\dfrac{x}{a}\right)^p+\displaystyle\left(\dfrac{y}{a}\right)^p' class='latex' />.</p>
<p>Jadi</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=1%26%2360%3B%5Cdisplaystyle%5Cleft%28%5Cdfrac%7Bx%7D%7Ba%7D%5Cright%29%5Ep%2B%5Cdisplaystyle%5Cleft%28%5Cdfrac%7By%7D%7Ba%7D%5Cright%29%5Ep&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='1&lt;\displaystyle\left(\dfrac{x}{a}\right)^p+\displaystyle\left(\dfrac{y}{a}\right)^p' title='1&lt;\displaystyle\left(\dfrac{x}{a}\right)^p+\displaystyle\left(\dfrac{y}{a}\right)^p' class='latex' />.</p>
<p>Kalikan kedua ruas dengan <img src='http://l.wordpress.com/latex.php?latex=a%5Ep&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^p' title='a^p' class='latex' /> dan substitusikan <img src='http://l.wordpress.com/latex.php?latex=a%3Dx%2By&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a=x+y' title='a=x+y' class='latex' />. Maka</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%28x%2By%29%5Ep%26%2360%3Bx%5Ep%2By%5Ep&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(x+y)^p&lt;x^p+y^p' title='(x+y)^p&lt;x^p+y^p' class='latex' />.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Persamaan kuadrat]]></title>
<link>http://artofmathematics.wordpress.com/2008/02/02/persamaan-kuadrat/</link>
<pubDate>Sat, 02 Feb 2008 15:58:17 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/02/02/persamaan-kuadrat/</guid>
<description><![CDATA[[Rusia 1989] Tentukan akar-akar dari persamaan , jika diketahui bahwa akar-akarnya adalah bilangan b]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[Rusia 1989] Tentukan akar-akar dari persamaan <img src='http://l.wordpress.com/latex.php?latex=x%5E2%2Bpx%2Bq%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+px+q=0' title='x^2+px+q=0' class='latex' />, jika diketahui bahwa akar-akarnya adalah bilangan bulat dan <img src='http://l.wordpress.com/latex.php?latex=p%2Bq%3D198&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p+q=198' title='p+q=198' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Perhatikan bahwa <img src='http://l.wordpress.com/latex.php?latex=p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p' title='p' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=q&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='q' title='q' class='latex' /> pasti bilangan bulat karena akar-akarnya adalah bilangan bulat. Substitusikan <img src='http://l.wordpress.com/latex.php?latex=q%3D198-p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='q=198-p' title='q=198-p' class='latex' /> ke persamaan tadi, menjadi: <img src='http://l.wordpress.com/latex.php?latex=x%5E2%2Bpx%2B198-p%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+px+198-p=0' title='x^2+px+198-p=0' class='latex' />.</p>
<p>Diskriminannya</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=D%3Dp%5E2-4%28198-p%29%3Dp%5E2%2B4p-792&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='D=p^2-4(198-p)=p^2+4p-792' title='D=p^2-4(198-p)=p^2+4p-792' class='latex' /></p>
<p>adalah bilangan kuadrat. Maka misalkan <img src='http://l.wordpress.com/latex.php?latex=D%3Dm%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='D=m^2' title='D=m^2' class='latex' />, sehingga</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=p%5E2%2B4p-792%3Dm%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p^2+4p-792=m^2' title='p^2+4p-792=m^2' class='latex' />,</p>
<p>atau</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%28p%2B2%29%5E2-m%5E2%3D796&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(p+2)^2-m^2=796' title='(p+2)^2-m^2=796' class='latex' />,</p>
<p>dan</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%28p%2B2-m%29%28p%2B2%2Bm%29%3D1%5Ccdot796%3D2%5Ccdot398%3D4%5Ccdot199&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='(p+2-m)(p+2+m)=1\cdot796=2\cdot398=4\cdot199' title='(p+2-m)(p+2+m)=1\cdot796=2\cdot398=4\cdot199' class='latex' /></p>
<p>Jika <img src='http://l.wordpress.com/latex.php?latex=p%2B2-m%3D1&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p+2-m=1' title='p+2-m=1' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=p%2B2%2Bm%3D796&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p+2+m=796' title='p+2+m=796' class='latex' />, maka <img src='http://l.wordpress.com/latex.php?latex=p&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p' title='p' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=m&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m' title='m' class='latex' /> bukan bilangan bulat. Begitu pula jika <img src='http://l.wordpress.com/latex.php?latex=p%2B2-m%3D4&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p+2-m=4' title='p+2-m=4' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=p%2B2%2Bm%3D199&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p+2+m=199' title='p+2+m=199' class='latex' />.</p>
<p>Jadi <img src='http://l.wordpress.com/latex.php?latex=p%2B2-m%3D2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p+2-m=2' title='p+2-m=2' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=p%2B2%2Bm%3D398&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p+2+m=398' title='p+2+m=398' class='latex' />, yang menyebabkan <img src='http://l.wordpress.com/latex.php?latex=p%3D198&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p=198' title='p=198' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=m%3D198&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='m=198' title='m=198' class='latex' />. Maka <img src='http://l.wordpress.com/latex.php?latex=q%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='q=0' title='q=0' class='latex' />. Maka, persamaan itu adalah <img src='http://l.wordpress.com/latex.php?latex=p%28x%29%3Dx%5E2%2B198x%3Dx%28x%2B198%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='p(x)=x^2+198x=x(x+198)' title='p(x)=x^2+198x=x(x+198)' class='latex' />, dan akar-akarnya adalah <img src='http://l.wordpress.com/latex.php?latex=x_1%3D0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_1=0' title='x_1=0' class='latex' /> dan <img src='http://l.wordpress.com/latex.php?latex=x_2%3D-198&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x_2=-198' title='x_2=-198' class='latex' />.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Pertidaksamaan tiga bilangan dengan hasil kali 2]]></title>
<link>http://artofmathematics.wordpress.com/2008/01/03/pertidaksamaan-tiga-bilangan-dengan-hasil-kali-2/</link>
<pubDate>Thu, 03 Jan 2008 10:28:56 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2008/01/03/pertidaksamaan-tiga-bilangan-dengan-hasil-kali-2/</guid>
<description><![CDATA[[olimpiade.org] , , dan adalah tiga bilangan real positif dengan hasil kali 2. Buktikan bahwa . Solu]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[olimpiade.org] <img src='http://l.wordpress.com/latex.php?latex=a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a' title='a' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=b&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='b' title='b' class='latex' />, dan <img src='http://l.wordpress.com/latex.php?latex=c&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='c' title='c' class='latex' /> adalah tiga bilangan real positif dengan hasil kali 2. Buktikan bahwa</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=a%5E3%2Bb%5E3%2Bc%5E3%5Cge+a%5Csqrt%7Bb%2Bc%7D%2Bb%5Csqrt%7Bc%2Ba%7D%2Bc%5Csqrt%7Ba%2Bb%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^3+b^3+c^3\ge a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}' title='a^3+b^3+c^3\ge a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Perhatikan bahwa</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=+%28x-y%29%5E2%28x%2By%29%5Cge0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt=' (x-y)^2(x+y)\ge0' title=' (x-y)^2(x+y)\ge0' class='latex' />.</p>
<p>Maka</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=x%5E3%2By%5E3-x%5E2y-xy%5E2%5Cge0&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^3+y^3-x^2y-xy^2\ge0' title='x^3+y^3-x^2y-xy^2\ge0' class='latex' />,</p>
<p>atau</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=x%5E3%2By%5E3%5Cge+xy%28x%2By%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^3+y^3\ge xy(x+y)' title='x^3+y^3\ge xy(x+y)' class='latex' />.</p>
<p>Sekarang,</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=a%5E3%2Bb%5E3%2Bc%5E3%3D%5Cdfrac%7B1%7D%7B2%7D%5Cdisplaystyle%5Cleft%28a%5E3%2B%5Cdfrac%7Bb%5E3%2Bc%5E3%7D%7B2%7D%5Cright%29%2B%5Cdfrac%7B1%7D%7B2%7D%5Cdisplaystyle%5Cleft%28b%5E3%2B%5Cdfrac%7Bc%5E3%2Ba%5E3%7D%7B2%7D%5Cright%29%2B%5Cdfrac%7B1%7D%7B2%7D%5Cdisplaystyle%5Cleft%28c%5E3%2B%5Cdfrac%7Ba%5E3%2Bb%5E3%7D%7B2%7D%5Cright%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^3+b^3+c^3=\dfrac{1}{2}\displaystyle\left(a^3+\dfrac{b^3+c^3}{2}\right)+\dfrac{1}{2}\displaystyle\left(b^3+\dfrac{c^3+a^3}{2}\right)+\dfrac{1}{2}\displaystyle\left(c^3+\dfrac{a^3+b^3}{2}\right)' title='a^3+b^3+c^3=\dfrac{1}{2}\displaystyle\left(a^3+\dfrac{b^3+c^3}{2}\right)+\dfrac{1}{2}\displaystyle\left(b^3+\dfrac{c^3+a^3}{2}\right)+\dfrac{1}{2}\displaystyle\left(c^3+\dfrac{a^3+b^3}{2}\right)' class='latex' />.</p>
<p>Substitusikan pertidaksamaan tadi, sehingga</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=a%5E3%2Bb%5E3%2Bc%5E3%5Cge%5Cdfrac%7B1%7D%7B2%7D%5Cdisplaystyle%5Cleft%28a%5E3%2B%5Cdfrac%7Bbc%28b%2Bc%29%7D%7B2%7D%5Cright%29%2B%5Cdfrac%7B1%7D%7B2%7D%5Cdisplaystyle%5Cleft%28b%5E3%2B%5Cdfrac%7Bca%28c%2Ba%29%7D%7B2%7D%5Cright%29%2B%5Cdfrac%7B1%7D%7B2%7D%5Cdisplaystyle%5Cleft%28c%5E3%2B%5Cdfrac%7Bab%28a%2Bb%29%7D%7B2%7D%5Cright%29&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^3+b^3+c^3\ge\dfrac{1}{2}\displaystyle\left(a^3+\dfrac{bc(b+c)}{2}\right)+\dfrac{1}{2}\displaystyle\left(b^3+\dfrac{ca(c+a)}{2}\right)+\dfrac{1}{2}\displaystyle\left(c^3+\dfrac{ab(a+b)}{2}\right)' title='a^3+b^3+c^3\ge\dfrac{1}{2}\displaystyle\left(a^3+\dfrac{bc(b+c)}{2}\right)+\dfrac{1}{2}\displaystyle\left(b^3+\dfrac{ca(c+a)}{2}\right)+\dfrac{1}{2}\displaystyle\left(c^3+\dfrac{ab(a+b)}{2}\right)' class='latex' />.</p>
<p>Dengan AM-GM, kita lanjutkan</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=a%5E3%2Bb%5E3%2Bc%5E3%5Cge%5Csqrt%7B%5Cdfrac%7Ba%5E3bc%28b%2Bc%29%7D%7B2%7D%7D%2B%5Csqrt%7B%5Cdfrac%7Bb%5E3ca%28c%2Ba%29%7D%7B2%7D%7D%2B%5Csqrt%7B%5Cdfrac%7Bc%5E3ab%28a%2Bb%29%7D%7B2%7D%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^3+b^3+c^3\ge\sqrt{\dfrac{a^3bc(b+c)}{2}}+\sqrt{\dfrac{b^3ca(c+a)}{2}}+\sqrt{\dfrac{c^3ab(a+b)}{2}}' title='a^3+b^3+c^3\ge\sqrt{\dfrac{a^3bc(b+c)}{2}}+\sqrt{\dfrac{b^3ca(c+a)}{2}}+\sqrt{\dfrac{c^3ab(a+b)}{2}}' class='latex' />.</p>
<p>Substitusikan nilai <img src='http://l.wordpress.com/latex.php?latex=abc%3D2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='abc=2' title='abc=2' class='latex' />, dan sederhanakan sehingga</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=a%5E3%2Bb%5E3%2Bc%5E3%5Cge+a%5Csqrt%7Bb%2Bc%7D%2Bb%5Csqrt%7Bc%2Ba%7D%2Bc%5Csqrt%7Ba%2Bb%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^3+b^3+c^3\ge a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}' title='a^3+b^3+c^3\ge a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}' class='latex' />.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Benteng di papan catur]]></title>
<link>http://artofmathematics.wordpress.com/2007/12/31/benteng-di-papan-catur/</link>
<pubDate>Mon, 31 Dec 2007 04:35:07 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2007/12/31/benteng-di-papan-catur/</guid>
<description><![CDATA[[Mathematics and Chess] Dalam papan catur, setiap kotak diberi bilangan 1 sampai 64, seperti gambar ]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[Mathematics and Chess] Dalam papan catur, setiap kotak diberi bilangan 1 sampai 64, seperti gambar ini:</p>
<p align="center"><img src="http://artofmathematics.wordpress.com/files/2008/03/untitled-123892.jpg" alt="papan catur bernomor" /></p>
<p>Delapan buah benteng ditempatkan, sehingga tidak ada yang dapat menyerang satu sama lain. Berapa jumlah dari bilangan-bilangan yang ditempati benteng-benteng itu? Bagaimana jika papan itu berukuran <img src='http://l.wordpress.com/latex.php?latex=n%5Ctimes+n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n\times n' title='n\times n' class='latex' /> dan terdapat <img src='http://l.wordpress.com/latex.php?latex=n&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n' title='n' class='latex' /> benteng?</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Misalkan <img src='http://l.wordpress.com/latex.php?latex=a_%7Bi%2Cj%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_{i,j}' title='a_{i,j}' class='latex' /> adalah bilangan pada baris ke <img src='http://l.wordpress.com/latex.php?latex=i&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='i' title='i' class='latex' /> dan kolom ke <img src='http://l.wordpress.com/latex.php?latex=j&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='j' title='j' class='latex' />. Maka, <img src='http://l.wordpress.com/latex.php?latex=a_%7Bi%2Cj%7D%3Dn%28i-1%29%2Bj&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a_{i,j}=n(i-1)+j' title='a_{i,j}=n(i-1)+j' class='latex' />. Dalam setiap baris dan setiap kolom hanya terdapat satu benteng. Maka jumlahnya</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dn%28i-1%29%2B%5Cdisplaystyle%5Csum_%7Bj%3D1%7D%5E%7Bn%7Dj%3Dn%5Ccdot%5Cdfrac%7Bn%28n-1%29%7D%7B2%7D%2B%5Cdfrac%7Bn%28n%2B1%29%7D%7B2%7D%3D%5Cdfrac%7Bn%5E3%2Bn%7D%7B2%7D&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\displaystyle\sum_{i=1}^{n}n(i-1)+\displaystyle\sum_{j=1}^{n}j=n\cdot\dfrac{n(n-1)}{2}+\dfrac{n(n+1)}{2}=\dfrac{n^3+n}{2}' title='\displaystyle\sum_{i=1}^{n}n(i-1)+\displaystyle\sum_{j=1}^{n}j=n\cdot\dfrac{n(n-1)}{2}+\dfrac{n(n+1)}{2}=\dfrac{n^3+n}{2}' class='latex' />.</p>
<p>Untuk papan catur <img src='http://l.wordpress.com/latex.php?latex=8%5Ctimes8&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='8\times8' title='8\times8' class='latex' />, substitusikan <img src='http://l.wordpress.com/latex.php?latex=n%3D8&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='n=8' title='n=8' class='latex' />, sehingga didapat jumlahnya 260.</p>
</div>]]></content:encoded>
</item>
<item>
<title><![CDATA[Pertidaksamaan]]></title>
<link>http://artofmathematics.wordpress.com/2007/12/05/pertidaksamaan-6/</link>
<pubDate>Wed, 05 Dec 2007 10:34:04 +0000</pubDate>
<dc:creator>Johan</dc:creator>
<guid>http://artofmathematics.wordpress.com/2007/12/05/pertidaksamaan-6/</guid>
<description><![CDATA[[Russian 2002] Jika , , adalah bilangan real positif dengan jumlah , buktikan bahwa . Solusi Dari pe]]></description>
<content:encoded><![CDATA[<div class='snap_preview'><p>[Russian 2002] Jika <img src='http://l.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=y&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y' title='y' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=z&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='z' title='z' class='latex' /> adalah bilangan real positif dengan jumlah <img src='http://l.wordpress.com/latex.php?latex=3&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='3' title='3' class='latex' />, buktikan bahwa</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7Bx%7D%2B%5Csqrt%7By%7D%2B%5Csqrt%7Bz%7D%5Cge+xy%2Byz%2Bzx&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{x}+\sqrt{y}+\sqrt{z}\ge xy+yz+zx' title='\sqrt{x}+\sqrt{y}+\sqrt{z}\ge xy+yz+zx' class='latex' />.</p>
<p><!--more Lihat Solusi --></p>
<p>Solusi<br />
Dari pertidaksamaan AM-GM,</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=a%5E2%2B%5Csqrt%7Ba%7D%2B%5Csqrt%7Ba%7D%5Cge3a&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='a^2+\sqrt{a}+\sqrt{a}\ge3a' title='a^2+\sqrt{a}+\sqrt{a}\ge3a' class='latex' />.</p>
<p>Substitusikan untuk <img src='http://l.wordpress.com/latex.php?latex=x&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x' title='x' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=y&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='y' title='y' class='latex' />, dan <img src='http://l.wordpress.com/latex.php?latex=z&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='z' title='z' class='latex' /> kemudian jumlahkan,</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=x%5E2%2By%5E2%2Bz%5E2%2B2%5Csqrt%7Bx%7D%2B2%5Csqrt%7By%7D%2B2%5Csqrt%7Bz%7D%5Cge3x%2B3y%2B3z&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+y^2+z^2+2\sqrt{x}+2\sqrt{y}+2\sqrt{z}\ge3x+3y+3z' title='x^2+y^2+z^2+2\sqrt{x}+2\sqrt{y}+2\sqrt{z}\ge3x+3y+3z' class='latex' />.</p>
<p align="left">Substitusikan <img src='http://l.wordpress.com/latex.php?latex=3%3Dx%2By%2Bz&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='3=x+y+z' title='3=x+y+z' class='latex' />,</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=x%5E2%2By%5E2%2Bz%5E2%2B2%5Csqrt%7Bx%7D%2B2%5Csqrt%7By%7D%2B2%5Csqrt%7Bz%7D%5Cge%28x%2By%2Bz%29%5E2&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='x^2+y^2+z^2+2\sqrt{x}+2\sqrt{y}+2\sqrt{z}\ge(x+y+z)^2' title='x^2+y^2+z^2+2\sqrt{x}+2\sqrt{y}+2\sqrt{z}\ge(x+y+z)^2' class='latex' /></p>
<p>Sederhanakan,</p>
<p align="center"><img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7Bx%7D%2B%5Csqrt%7By%7D%2B%5Csqrt%7Bz%7D%5Cge+xy%2Byz%2Bzx&#038;bg=ffffff&#038;fg=000000&#038;s=0' alt='\sqrt{x}+\sqrt{y}+\sqrt{z}\ge xy+yz+zx' title='\sqrt{x}+\sqrt{y}+\sqrt{z}\ge xy+yz+zx' class='latex' />.</p>
</div>]]></content:encoded>
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