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
是恒等式,则A-B=______. B x+1
答
∵
+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.