数学问题:若(x-5)/(x^2-1)=A/(x+1)+B/(x-1)是恒等式,求A.B的值.

来源：百度知道 编辑：UC知道 时间：2024/05/26 17:27:36
写明过程,谢谢了~

'先进行通分
(x-5)/(x^2-1)=A/(x+1)+B/(x-1)＝A(x-1)+B(X+1)/(x^2-1)
’因为是恒等式，所以

(x-5)=A(x-1)+B(X+1)=(A+B)X+(B-A)
所以
A＋B＝1
B－A＝5
结果为
A＝－2，B＝3

(x-5)/(x^2-1)=A/(x+1)+B/(x-1)
(x-5)/(x^2-1)=(x-5)/[(x+1)(x-1)]
A/(x+1)+B/(x-1)=[A(x-1)+B(x+1)]/[(x+1)(x-1)]=[(A+B)x+(A-B)]/[(x+1)(x-1)]
(x-5)/(x^2-1)=A/(x+1)+B/(x-1)是恒等式
所以
A+B=1
A-B=-5
A=-2
B=3