问题: 已知
已知(x²+x-3)/[(x-1)(x-2)(x-3)]=A/(x-1)+B/(x-2)+C/(x-3),求A，B，C的值．
解答:
A/(x-1)+B/(x-2)+C/(x-3)
=[A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)]/[(x-1)(x-2)(x-3)]
=[A(x^2-5x+6)+B(x^2-4x+3)+C(x^2-3x+2)]/[(x-1)(x-2)(x-3)]
=[(A+B+C)x^2-(5A+4B+3C)x+(6A+3B+2C)]/(x²+x-3)/[(x-1)(x-2)(x-3)]
=(x²+x-3)/[(x-1)(x-2)(x-3)]
所以A+B+C=1，-(5A+4B+3C)=1，6A+3B+2C=-3
解得A=-1/2，B=-3，C=9/2
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