若4/(x^2-1)=A/(x+1)+B/(x-1)是恒等式,则A=?B=?

若4/(x^2-1)=A/(x+1)+B/(x-1)是恒等式,则A=?B=?
若4/(x^2-1)=A/(x+1)+B/(x-1)是恒等式,则A=?B=?

若4/(x^2-1)=A/(x+1)+B/(x-1)是恒等式,则A=?B=?
右边=A/(x+1)+B/(x-1)
=[A(x-1)+B(x+1)]/[(x+1)(x-1)]
=[Ax+Bx+B-A]/(x^2-1)
=[(A+B)x+B-A]/(x^2-1)
因为左边=4/(x^2-1)且等式恒成立,分母已经相同,那么分子也要相同
所以(A+B)x+B-A=4
该式对任意x都成立的化,就有A+B=0,B-A=4
解方程,就得A=-2,B=2

4=A(X-1)+B(X+1)
4=(A+B)X+B-A
A+B=0
B-A=4
A=-2
B=2