若A/(x+3)+B/(x-2)=(3x+4)/(x^2+x-6),求A,B.

问题描述:

若A/(x+3)+B/(x-2)=(3x+4)/(x^2+x-6),求A,B.

等式左边通分,得A/(x+3)+B/(x-2)=(3x+4)/(x^2+x-6)[A(x-2)+B(x+3)]/(x+3)(x-2)=(3x+4)/(x^2+x-6)[(A+B)x+3B-2A]/(x^2+x-6)=(3x+4)/(x^2+x-6)上式分母相等,所以分子对照,得A+B=33B-2A=4解这个方程组,得A=1B=2...