1、根据题意
(3x+1)/(x²-x-6)=A/(x-3)+B/(x+2)
(3x+1)/(x²-x-6)=[A(x+2)+B(x-3)]/[(x-3)(x+2)]
(3x+1)/(x²-x-6)=(Ax+2A+Bx-3B)/(x²-x-6)
(3x+1)/(x²-x-6)=[(A+B)x+(2A-3B)]/(x²-x-6)
A+B=3(1)
2A-3B=1(2)
(1)×3+(2)
5A=10
A=2
B=3-A=1
A=2
B=1
2、y=|x-1|/(x-1)-|x-2|/(x-2)+|x²-3x+2|/(3x-x²-2)
=|x-1|/(x-1)-|x-2|/(x-2)-|(x-2)(x-1)|/(x²-3x+2)
=|x-1|/(x-1)-|x-2|/(x-2)-|(x-1)(x-2)|/[(x-1)(x-2)]
零点法
讨论
(1)x