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※ 引述《shanewang43 (WSG)》之銘言: : f(x)除以(x-2)^2為13x-17 : 除以(x-1)^2為3x-1 : 求除以[(x-1)^2][(x-2)^2]的餘式 : 想不出來 跪求想法 令 f(x) = [(x-1)^2][(x-2)^2]p(x) + a(x-1)(x-2)^2 + b(x-2)^2 + 13x - 17 又 f(1) = 2 = b - 4 => b = 6 => f(x) = [(x-1)^2][(x-2)^2]p(x) + a(x-1)(x-2)^2 + 6(x-2)^2 + 13x - 17 = [(x-1)^2][(x-2)^2]p(x) + a(x-1)[(x-1)(x-3) + 1] + 6[(x-1)(x-1) - 2x + 3] + 13x - 17 = [(x-1)^2][(x-2)^2]p(x) + a(x-3)(x-1)^2 + 6(x-1)^2 + (1 + a)x + 1 - a = G(x)(x-1)^2 + (1 + a)x + 1 - a => 1 + a = 3, 1 - a = -1 得 a = 2, 所求為 2(x-1)(x-2)^2 + 6(x-2)^2 + 13x - 17 -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 118.169.47.104 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1538149881.A.B36.html
shanewang43 : 感謝 有啟發到我了 10/01 10:51