题目

已知:a^2-4a-5=0,试求a^2-5a+6分之a^3-2a^2-9a+21减a^2-3a+2分之a^3-a^2-4a+1
我主要是想知道分子怎么化简，这里你所说的我都知道，

提问时间：2021-12-25

答案

已知a^2-4a-5=0
解法：利用配项法将高次项消去,然后再通分.
(a^3-2a^2-9a+21)/(a^2-5a+6) - (a^3-a^2-4a+1)/(a^2-3a+2)
= [(a^3-4a^2-5a)+2a^2-4a+21]/(a^2-5a+6) - [(a^3-4a^2-5a)+3a^2+a+1]/(a^2-3a+2)
= [(a^2-4a-5)a+2a^2-4a+21]/(a^2-5a+6) - [(a^2-4a-5)a+3a^2+a+1]/(a^2-3a+2)
= (2a^2-4a+21)/(a^2-5a+6) - (3a^2+a+1)/(a^2-3a+2)
= [(2a^2-8a-10)+4a+31]/(a^2-5a+6) - [(3a^2-12a-15)+13a+16]/(a^2-3a+2)
= [2(a^2-4a-5)+4a+31]/(a^2-5a+6) - [3(a^2-4a-5)+13a+16]/(a^2-3a+2)
= (4a+31)/(a^2-5a+6) - (13a+16)/(a^2-3a+2)
= (4a+31)/[(a-2)(a-3)] - (13a+16)/[(a-1)(a-2)]
= [(4a+31)(a-1)-(13a+16)(a-3)]/[(a-1)(a-2)(a-3)]
= (-9a^2+50a-79)/[(a-1)(a-2)(a-3)]
= [(-9a^2+36a+45)+14a-124]/[(a-1)(a-2)(a-3)]
= [-9(a^2-4a-5)+14a-124]/[(a-1)(a-2)(a-3)]
= (-14a-124)/[(a-1)(a-2)(a-3)]
= -2(7a+62)/[(a-1)(a-2)(a-3)]