3x-4⼀(x-1)(x-2)=A⼀(x-1)+B⼀(x-2)

“/”表示分数线 要有具体的解答步骤谢谢
2025-05-12 00:08:54
推荐回答(2个)
回答1:

3x-4/(x-1)(x-2)=A/(x-1)+B/(x-2)
(3x-4)/(x-1)(x-2)=[A(x-2)+B(x-1)]/(x-1)(x-2)
(3x-4)/(x-1)(x-2)=[(A+B)x-(2A+B)]/(x-1)(x-2)
两边恒等
A+B=3
2A+B=4
A=1
B=2

回答2:

A/(x-1)+B/(x-2)=[(A+B)x-(2A+B)]/(x-1)(x-2)

所以A+B=3,2A+B=4
A=1,B=2

A=1 B=2时,x也不能取1,2,否则分母就等于0,没意义了