(x-3)/(x^2-3x+2)=A/(x-1)-B/(x-2)
(x-3)/[(x-1)(x-2)]=A(x-2)/[(x-1)(x-2)-B(x-1)/[(x-1)(x-2)]
(x-3)/[(x-1)(x-2)]=[A(x-2)-B(x-2)]/[(x-1)(x-2)]
当x≠1、x≠2时,有:
x-3=A(x-2)-B(x-1)
x-3=Ax-2A-Bx+B
x-3=(A-B)x-2A+B
所以:
A-B=1…………………①
2A-B=3………………②
解上述二元一次方程组,
②-①得:A=2,
倒入①得:2-B=1
解得:B=1
即:A=2、B=1