1/(x+2)(x+1)(1-x)怎么分解成和的形式我知道带入abc求解,请给个具体过程,

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1/(x+2)(x+1)(1-x)怎么分解成和的形式我知道带入abc求解,请给个具体过程,1/(x+2)(x+1)(1-x)怎么分解成和的形式我知道带入abc求解,请给个具体过程,1/(x+2)(x+

1/(x+2)(x+1)(1-x)怎么分解成和的形式我知道带入abc求解,请给个具体过程,
1/(x+2)(x+1)(1-x)怎么分解成和的形式
我知道带入abc求解,请给个具体过程,

1/(x+2)(x+1)(1-x)怎么分解成和的形式我知道带入abc求解,请给个具体过程,
1/(x+2)(x+1)(1-x)
=1/(x+2) * 1/(x+1)(1-x)
=1/(x+2) * 1/2[1/(x+1) + 1/(1-x)]
=1/2 * [1/(x+2)(x+1) + 1/(x+2)(1-x)]
=1/2 * [1/(x+1)-1/(x+2) + 1/3*( 1/(x+2)+1/(1-x) )]
=1/2 * [1/(x+1)-2/3(x+2) + 1/3(1-x) )]
=1/2(x+1) -1/3(x+2) +1/6(1-x)
我觉得这么算不容易乱,当然设abc也可以
设原式=a/(x+2)+b/(x+1)+c/(1-x)
a/(x+2)+b/(x+1)+c/(1-x)=[a(x+1)(1-x)+b(x+2)(1-x)+c(x+2)(x+1)]/(x+1)(1-x)(x+2) =[(-a-b+c)x^2+(-b+3c)x+(a+2b+2c)]/(x+1)(1-x)(x+2)
x和x²项系数为0,常数项为1,则
-a-b+c=0
-b+3c=0
a+2b+2c=1,
a=-1/3,b=1/2,c=1/6

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