若f(x)÷x^2-4x+3,餘式為-2x+5,又當f(x)÷x^2-3x+2,餘式為kx+8,
則當 [f(x)]^2007÷x-2,,餘式/數是多少??
[數學 Maths] 十萬火急!!!----餘式定理
2010-09-22 3:01 am
回答 (1)
2010-09-22 4:30 am
✔ 最佳答案
f(x) ÷ (x^2-4x+3 = (x - 1)(x - 3) ) ,餘式為 - 2x + 5 :故 f(1) = - 2(1) + 5 = 3 ,f(x) ÷ (x^2-3x+2 = (x - 1)(x - 2) ),餘式為 kx + 8 :故 f(1) = k(1) + 8 = 3 ,k = - 5故 f(2) = k(2) + 8 = -5(2) + 8 = - 2當 [f(x)]^2007 ÷ (x-2) , 餘式/數是 [f(2)]^2007 = (- 2)^2007 = - 2^2007
收錄日期: 2021-04-21 22:14:51
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https://hk.answers.yahoo.com/question/index?qid=20100921000051KK01051