Re: [中學] 因式分解

看板Math作者Tiderus (修煉人生)時間8年前 (2017/06/02 13:18)推噓1(1推 0噓 4→)

留言5則, 2人參與討論串58/78 (看更多)

※ 引述《tcbt32 (唐)》之銘言： : https://imgur.com/gallery/a54Ti : 麻煩各位大大 (1) 原= (x^4 + 4y^4) + (x^3 y + 2xy^3) + 2 x^2 *y^2 = (x^2 + 2y^2)^2 + xy(x^2 + 2y^2) - 2 x^2 *y^2 = [(x^2 + 2y^2) + 2xy][(x^2 + 2y^2) - xy] (2) 原= (x^4 + 9) + (x^3 + 3x) + 4x^2 = (x^2 + 3)^2 + x(x^2 + 3) - 2x^2 = [(x^2 + 3) + 2x][(x^2 + 3) - x] -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 123.240.120.14 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1496380697.A.6AE.html

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yyc2008

06/02 13:21, , 1^F

06/02 13:21, 1^F

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yyc2008

06/02 13:21, , 2^F

06/02 13:21, 2^F

應該是為了弄出常數吧 A = x + k/x --> A^2 = x^2 + (k/x)^2 + 2k ^^^^常數 (1) B = (x/y) + 2/(x/y) --> B^2 = (x/y)^2 + 2^2/(x/y)^2 + 4 原= (x^2 y^2){[(x/y)^2 + 4(y/x)^2) + [(x/y) + 2(y/x)] + 2} = (x^2 y^2){[B^2 - 4] + B + 2} = (x^2 y^2)(B^2 + B -2) = (x^2 y^2)(B + 2)(B - 1) = [xy(B + 2)][xy(B - 1)] = [(x^2 + 2y^2) + 2xy][(x^2 + 2y^2) - xy] ※ 編輯: Tiderus (123.240.120.14), 06/02/2017 13:33:32

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yyc2008

06/02 13:38, , 3^F

06/02 13:38, 3^F

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yyc2008

06/02 13:38, , 4^F

06/02 13:38, 4^F

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Vulpix

06/03 00:04, , 5^F

06/03 00:04, 5^F