Question

若X的平方减1分之4等于X+1分之A,加上X-1分之B是恒等式,则A.B是多少?.

Answer 1

4/(x²-1)=A/(x+1)+B/(x-1)

=[A(x-1)+B(x+1)]/(x²-1)

=[(A+B)x+B-A]/(x²-1)是恒等式

则有

A+B=0

B-A=4

A=-2,B=2