{"id":2621,"date":"2022-05-01T22:18:03","date_gmt":"2022-05-01T20:18:03","guid":{"rendered":"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/binomicka-veta-a-pascaluv-trojuhelnik\/"},"modified":"2022-05-02T14:49:45","modified_gmt":"2022-05-02T12:49:45","slug":"binomicka-veta-a-pascaluv-trojuhelnik","status":"publish","type":"lesson","link":"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/binomicka-veta-a-pascaluv-trojuhelnik\/","title":{"rendered":"Binomick\u00e1 v\u011bta a Pascal\u016fv troj\u00faheln\u00edk"},"content":{"rendered":"<p>Existuj\u00ed vzore\u010dky, kter\u00e9 si pamatujeme a um\u00edme je zpam\u011bti. Ale n\u011bkdy se stane, \u017ee i ty nejzn\u00e1m\u011bj\u0161\u00ed vzore\u010dky n\u00e1m vypadnou z hlavy. Proto je praktick\u00e9 um\u011bt si vzorce odvodit a t\u00edm je l\u00e9pe pochopit. Proto se pod\u00edv\u00e1me na to, jak se daj\u00ed odvodit vzore\u010dky pro v\u00fdpo\u010det: <strong><em>(a+b)\u00b2=a\u00b2+2ab+b\u00b2<\/em><\/strong><\/p>\n<ul>\n<li>K odvozen\u00ed v\u00fdpo\u010dtu pro <strong><em>(a+b)\u00b3<\/em><\/strong>, ale t\u0159eba i na \u010dtvrtou \u010di p\u00e1tou n\u00e1m m\u016f\u017ee poslou\u017eit <strong>Pascal\u016fv troj\u00faheln\u00edk<\/strong>, kter\u00fd m\u00e1 ale i celou \u0159adu dal\u0161\u00edch vyu\u017eit\u00ed. A proto se rozhodn\u011b vyplat\u00ed jej zn\u00e1t.<\/li>\n<li>Pascal\u016fv troj\u00faheln\u00edk je uskupen\u00ed \u010d\u00edsel, do troj\u00faheln\u00edku a vypad\u00e1 n\u00e1sledovn\u011b:<\/li>\n<\/ul>\n<p>$$\\begin{matrix}<br \/>\n&amp;&amp;&amp;&amp;&amp;1\\\\<br \/>\n&amp;&amp;&amp;&amp;1&amp;&amp;1\\\\<br \/>\n&amp;&amp;&amp;1&amp;&amp;2&amp;&amp;1\\\\<br \/>\n&amp;&amp;1&amp;&amp;3&amp;&amp;3&amp;&amp;1\\\\<br \/>\n&amp;1&amp;&amp;4&amp;&amp;6&amp;&amp;4&amp;&amp;1\\\\<br \/>\n1&amp;&amp;5&amp;&amp;10&amp;&amp;10&amp;&amp;5&amp;&amp;1<br \/>\n\\end{matrix}$$<\/p>\n<ul>\n<li>S vypisov\u00e1n\u00edm Pascalova troj\u00faheln\u00edku bychom mohli pokra\u010dovat d\u00e1l. Jakmile pochop\u00edte, jak Pascal\u016fv troj\u00faheln\u00edk funguje, budete schopni pokra\u010dovat v jeho psan\u00ed sami.<\/li>\n<li>P\u0159i bli\u017e\u0161\u00edm pohledu na Pascal\u016fv troj\u00faheln\u00edk toti\u017e zjist\u00edme, \u017ee <strong>\u010d\u00edsla v n\u011bm nejsou uspo\u0159\u00e1d\u00e1na n\u00e1hodn\u011b<\/strong>, ale maj\u00ed jasn\u011b dan\u00e1 m\u00edsta.<\/li>\n<li><strong>Ka\u017ed\u00e9 \u010d\u00edslo je sou\u010dtem \u010d\u00edsel nad n\u00edm.<\/strong> To je dob\u0159e patrn\u00e9 ze t\u0159et\u00edho \u0159\u00e1dku, ve kter\u00e9m se nach\u00e1z\u00ed \u010d\u00edsla <em><strong>1<\/strong><\/em>, <em><strong>2<\/strong><\/em> a <em><strong>1<\/strong><\/em>. Nad ob\u011bma jedni\u010dkami se v\u017edy nach\u00e1z\u00ed 1 pouze z jedn\u00e9 strany <strong><em>(0+1=1)<\/em><\/strong>. Kdy\u017e se pod\u00edv\u00e1me nad \u010d\u00edslo<em><strong> 2<\/strong><\/em>, zjist\u00edme \u017ee v \u0159\u00e1dku nad n\u00edm se vlevo od n\u011bj nach\u00e1z\u00ed \u010d\u00edslo <em><strong>1<\/strong><\/em> a vpravo \u010d\u00edslo <strong><em>2<\/em><\/strong> a <em><strong>1+1=2<\/strong><\/em>. I p\u0159i pohledu na dal\u0161\u00ed \u0159\u00e1dek je z\u0159ejm\u00e9, jak se v n\u011bm objevila \u010d\u00edsla <strong><em>1<\/em><\/strong>, <em><strong>3<\/strong><\/em>, <em><strong>3<\/strong><\/em>, <strong><em>1<\/em><\/strong>. Op\u011bt plat\u00ed, \u017ee: <em><strong>0+1=1<\/strong><\/em>, <strong><em>1+2=3<\/em><\/strong>,<em><strong> 2+1=3<\/strong><\/em>, <em><strong>1+0=1<\/strong><\/em>. Takto bychom mohli pokra\u010dovat pro ka\u017ed\u00fd \u0159\u00e1dek.<\/li>\n<li>Dost\u00e1v\u00e1me se tedy k ot\u00e1zce, jak Pascal\u016fv troj\u00faheln\u00edk vyu\u017e\u00edt v praxi?<\/li>\n<li>Jedn\u00edm ze zp\u016fsob\u016f vyu\u017eit\u00ed je<strong> Binomick\u00e1 v\u011bta<\/strong>. Poj\u010fme si ji uk\u00e1zat v praxi.<\/li>\n<li>Pot\u0159ebujete vytvo\u0159it vzore\u010dek <strong><em>(a+b)\u00b3<\/em><\/strong>, ale nev\u00edte si rady? \u0158e\u0161en\u00ed je snadn\u00e9: pod\u00edvejte se na<strong> 4. \u0159\u00e1dek Pascalova troj\u00faheln\u00edku<\/strong>, kter\u00fd se skl\u00e1d\u00e1 ze 4 \u010d\u00edsel. \u010c\u00edsla v tomto \u0159\u00e1dku jsou: <strong>1,3,3,1.\u00a0<\/strong><\/li>\n<li>V\u00fdpo\u010det vzorce <em><strong>(a+b)\u00b3<\/strong> <\/em>vypad\u00e1 n\u00e1sledovn\u011b:<\/li>\n<\/ul>\n<p>$${(a+b)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$$<\/p>\n<p style=\"text-align: center\">Pokud bychom tento z\u00e1pis nepatrn\u011b upravili a dopsali do n\u011bj i jedni\u010dky p\u0159ed <strong><em>a<\/em><\/strong> a <em><strong>b<\/strong><\/em>, dostali bychom toto:<\/p>\n<p>$${(a+b)}^{3}={1a}^{3}+3{a}^{2}b+3a{b}^{2}+{1b}^{3}$$<\/p>\n<p style=\"text-align: center\">Z takov\u00e9ho z\u00e1pisu je ji\u017e l\u00e9pe patrn\u00e9, jak m\u016f\u017eeme vyu\u017e\u00edt Pascal\u016fv troj\u00faheln\u00edk pro binomick\u00fd rozvoj.<\/p>\n<ul>\n<li>P\u0159i odvozov\u00e1n\u00ed binomick\u00e9ho rozvoje nesm\u00edme zapom\u00ednat na v\u0161echny \u010dleny. Je-li z\u00e1rovka (a+b) umocn\u011bna na t\u0159et\u00ed, v rozvoji se objev\u00ed \u010dleny umocn\u011bn\u00e9 na nultou, prvou, druhou i t\u0159et\u00ed. (viz p\u0159\u00edklad, kter\u00fd jsme odvodili z Pascalova troj\u00faheln\u00edku).<\/li>\n<\/ul>\n","protected":false},"comment_status":"closed","ping_status":"closed","template":"","class_list":["post-2621","lesson","type-lesson","status-publish","hentry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.5 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Binomick\u00e1 v\u011bta a Pascal\u016fv troj\u00faheln\u00edk | Studijni-svet.cz<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/binomicka-veta-a-pascaluv-trojuhelnik\/\" \/>\n<meta property=\"og:locale\" content=\"cs_CZ\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Binomick\u00e1 v\u011bta a Pascal\u016fv troj\u00faheln\u00edk | Studijni-svet.cz\" \/>\n<meta property=\"og:description\" content=\"Existuj\u00ed vzore\u010dky, kter\u00e9 si pamatujeme a um\u00edme je zpam\u011bti. Ale n\u011bkdy se stane, \u017ee i ty nejzn\u00e1m\u011bj\u0161\u00ed vzore\u010dky n\u00e1m vypadnou z hlavy. Proto je praktick\u00e9 um\u011bt si vzorce odvodit a t\u00edm je l\u00e9pe pochopit. Proto se pod\u00edv\u00e1me na to, jak se daj\u00ed odvodit vzore\u010dky pro v\u00fdpo\u010det: (a+b)\u00b2=a\u00b2+2ab+b\u00b2 K odvozen\u00ed v\u00fdpo\u010dtu pro (a+b)\u00b3, ale t\u0159eba i ... 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