{"id":194,"date":"2011-12-02T23:06:45","date_gmt":"2011-12-02T14:06:45","guid":{"rendered":"http:\/\/roundown.main.jp\/nyushi\/?p=194"},"modified":"2021-09-10T07:48:53","modified_gmt":"2021-09-09T22:48:53","slug":"osr200903","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/osr200903\/","title":{"rendered":"\u962a\u5927\u7406\u7cfb2009\uff1a\u7b2c3\u554f"},"content":{"rendered":"
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\\(\\alpha\\) \u3092 \\(2\\) \u6b21\u65b9\u7a0b\u5f0f \\(x^2-2x-1 = 0\\) \u306e\u89e3\u3068\u3059\u308b\u3068\u304d, \\(( a+5 \\alpha )( b+5c \\alpha )=1\\) \u3092\u307f\u305f\u3059\u6574\u6570\u306e\u7d44 \\((a, b, c)\\) \u3092\u3059\u3079\u3066\u6c42\u3081\u3088.\r\n\u305f\u3060\u3057, \u5fc5\u8981\u306a\u3089\u3070 \\(\\sqrt{2}\\) \u304c\u7121\u7406\u6570\u3067\u3042\u308b\u3053\u3068\u306f\u8a3c\u660e\u305b\u305a\u306b\u7528\u3044\u3066\u3088\u3044.<\/p>\r\n

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\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n

\u6761\u4ef6\u3088\u308a\r\n\\[\r\n\\alpha = 1 \\pm\\sqrt{2} , \\quad \\alpha^2 =2\\alpha +1\n\\]\r\n\u4e0e\u5f0f\u3088\u308a\r\n\\[\\begin{align}\r\nab+5( ac+b ) \\alpha +25c \\alpha^2 & =1 \\\\\r\nab+5( ac+b ) \\alpha +25c ( 2 \\alpha +1 ) -1 & =0 \\\\\r\n\\text{\u2234} \\quad \\underline{ab+25c-1} _ {\\text{[A]}} +5 \\alpha \\underline{(ac+b+10c)} _ {\\text{[B]}} & = 0\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\\begin{align}\r\n\\text{[A]} +5 \\left( 1 \\pm\\sqrt{2} \\right) \\text{[B]} & = 0 \\\\\r\n\\text{\u2234} \\quad \\left( \\text{[A]} +5 \\text{[B]} \\right) \\pm 5 \\sqrt{2} \\text{[B]} & = 0\n\\end{align}\\]\r\n\u3053\u308c\u304c\u6210\u7acb\u3059\u308b\u306e\u306f, \\(a , b , c\\) \u306f\u6574\u6570\u3067, \\(\\sqrt{2}\\) \u306f\u7121\u7406\u6570\u306a\u306e\u3067\r\n\\[\r\n\\left\\{ \\begin{array}{l} \\text{[A]} +5 \\text{[B]} =0 \\\\ \\text{[B]} =0 \\end{array} \\right.\n\\]\r\n\u3059\u306a\u308f\u3061, \\(\\text{[A]} =\\text{[B]} =0\\) \u306e\u3068\u304d\u3067\u3042\u308b.
\r\n\u3064\u307e\u308a\r\n\\[\r\n\\left\\{ \\begin{array}{ll} ab+25c-1 =0 & \\ \\text{...[1]} \\\\ ac+b+10c =0 & \\ \\text{...[2]} \\end{array} \\right.\n\\]\r\n[2] \u3088\u308a\r\n\\[\r\nb=-(a+10)c\n\\]\r\n[1] \u306b\u4ee3\u5165\u3057\u3066\r\n\\[\\begin{align}\r\n-ac(a+10) +25c-1 & =0 \\\\\r\n\\text{\u2234} \\quad c \\left( a^2+10a-25 \\right) & = -1\n\\end{align}\\]\r\n\\(c , a^2+10a-25\\) \u306f\u3068\u3082\u306b\u6574\u6570\u306a\u306e\u3067<\/p>\r\n

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  1. 1*<\/strong>\u3000\\(c =-1\\) , \\(a^2+10a-25 =1\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\na^2+10a-26 & =0 \\\\\r\n\\text{\u2234} \\quad a =-5 \\pm \\sqrt{25+26} & = -5 \\pm\\sqrt{51}\n\\end{align}\\]\r\n\u3053\u308c\u306f\u6574\u6570\u3067\u306a\u3044\u306e\u3067, \u4e0d\u9069.<\/p><\/li>\r\n

  2. 2*<\/strong>\u3000\\(c =1\\) , \\(a^2+10a-25 =-1\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\na^2+10a-24 & = 0 \\\\\r\n(a-2)(a+12) & =0 \\\\\r\n\\text{\u2234} \\quad a & =2 , -12\n\\end{align}\\]\r\n