Question

将分式f(x)=2x＾3-3x＾2-6x+17/x＾2-x-2表示成整式与部分分式的和：f(x)=Q(x)+A/x+a+B/x+b 求过程,谢谢

Answer 1

f(x)=(2x³-3x²-6x+17)/(x²-x-2)
=(2x³-2x²-4x-x²+x+2-3x+15)/(x²-x-2)
=[2x(x²-x-2)-(x²-x-2)-3(x-5)]/(x²-x-2)
=[(2x-1)(x²-x-2)-3(x-5)]/(x²-x-2)
=(2x-1)(x²-x-2)/(x²-x-2)-3(x-5)/(x²-x-2)
=(2x-1)-3(x-5)/(x²-x-2)
=(2x-1)-(3x-15)/[(x-2)(x+1)]

设(3x-15)/[(x-2)(x+1)]=A/(x-2)+B/(x+1)
A/(x-2)+B/(x+1)=[(A+B)x+(A-2B)]/[(x-2)(x+1)]
所以，A+B=3，A-2B=-15
解得，A=-3，B=6
所以，(3x-15)/[(x-2)(x+1)]=6/(x+1)-3/(x-2)

所以，f(x)=(2x³-3x²-6x+17)/(x²-x-2)
=(2x-1)+3/(x-2)-6/(x+1)