已知4/x2−1=A/x−1+B/x+1是恒等式,则A-B=_.

问题描述:

已知

4
x2−1
A
x−1
+
B
x+1
是恒等式,则A-B=______.

A
x−1
+
B
x+1

=
A(x+1)+B(x−1)
x2−1

=
(A+B)x+(A−B)
x2−1

=
4
x2−1

A+B=0
A−B=4

解得A=2,B=-2,
∴A-B=4.
故答案为4.