已知:(4x+1)/(x-2)(x-5)=A/(x-5)+B/(x-2),则A= ,B=
问题描述:
已知:(4x+1)/(x-2)(x-5)=A/(x-5)+B/(x-2),则A= ,B=
已知:(4x+1)/(x-2)(x-5)=A/(x-5)+B/(x-2),则A= ,B=
答
(4x+1)/(x-2)(x-5)=A/(x-2)+B/(x-5)
(4x+1)/(x-2)(x-5)=[A(x-5)+B(x-2)]/(x-2)(x-5)
4x+1=(A+B)x-5A-2B
因为等式总能成立
所以
A+B=4,
-5A-2B=1
解这个方程组
得到
A=-3,
B=7