若60/(x+1)(x+2)(x+3)=a/x+1+b/x-2+c/x+3其中a、b、c为常数求a+b+c的值

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答案和解析

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+b+c)x^2+(5a+4b+3c)x+6a+3b+2c](x+3)]/(x+1)(x+2)(x+3)=60/(x+1)(x+2)(x+3)
∴a+b+c=0

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