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=

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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=已知：(4x+1)/(x-2

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=
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=

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=
(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

已知：(4x+1)/[(x-2)(x-5)]=A/(x-5)+B/(x-2)，?
A/(x-5)+B/(x-2)
=[A(x-2)+B(X-5)]/[(x-2)(x-5)]
=[(A+B)x+(-2A-5B)]/[(x-5)(x-2)]
A+B=4
-2A-5B=1
解二元一次方程组得:
A=7,B=-3.

4x+1/（x-2）（x-5）=A/x-5+B/x-2
4x+1/（x-2）（x-5）=(A/(x-5) * (x-2)(x-5)+(B/(x-2) * (x-2)(x-5)/(x-2)(x-5)
4x+1/(x-2)(x-5)=A(x-2)+B(x-5)/(x-2)(x-5)
4x+1=A(x-2)+B(x-5)
4x+1=Ax-2A+Bx-5B
4x+1=(A+B)x+(-2A-5B)
A+B=4
-2A-5B=1
A=7
B=-3

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