(6x+5)/(x2-x-12)=A/(x-4)+B/(X+3).则A+B=

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(6x+5)/(x2-x-12)=A/(x-4)+B/(X+3).则A+B=
(6x+5)/(x2-x-12)=A/(x-4)+B/(X+3).则A+B=

(6x+5)/(x2-x-12)=A/(x-4)+B/(X+3).则A+B=
由题意:
A/(x-4)+B/(X+3).
=[A(x+3) +B(x-4)]/(x-4)(x+3)
=[(A+B)x+(3A-4B)]/(x^2-x-12)
=(6x+5)/(x2-x-12)
对照系数可以得出:
A+B=6
3A-4B=5
解得:A=29/7,B=13/7

右边分子分母有理化,得6x+5=A(x+3)+B(x-4),化简可得6x+5=Ax+3A+Bx-4B,得A+B=6,3A-4B=5,

右式通分,分母与左边的一样
看分子:
6x+5=Ax+3A+Bx-4B
根据,同幂次未知数前系数相同

A+B=6
3A-4B=5

方程的右边可化为[A(X+3)+B(X-4)]/(X-4)(X+3)=[(A+B)X+3A-4B]/X2-X-12
对比等式两边可得A+B=6

二楼回答不错

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