Question

已知1/(x-1)(x+2)=a/(x-1)+b/(x+2),求a , b 的值？

1/(x+1)(x-2)=A/x+1–B/x-2
用对一类特殊分式的探究方法求A、B

Answer 1

1/[(x-1)(x+2)]=a/(x-1)+b/(x+2),
去分母得1=a(x+2)+b(x-1)
...................=(a+b)x+2a-b,
比较系数得a+b=0,
..................2a-b=1,
解得a=1/3,b=-1/3.