Question

若4/(x^2-1)=A/(x+1)+B/(x-1)是恒等式,则A=?B=?A+B为啥等于0

Answer 1

解答:

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

A+B=0,是在恒等式(A+B)x+B-A=4两边比较系数而得的