x/(x+1)(x+2)(x+3) 有理真分式化为部分分式
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x/(x+1)(x+2)(x+3) 有理真分式化为部分分式
![x/(x+1)(x+2)(x+3) 有理真分式化为部分分式](/uploads/image/z/17743715-35-5.jpg?t=x%2F%28x%2B1%29%28x%2B2%29%28x%2B3%29+%E6%9C%89%E7%90%86%E7%9C%9F%E5%88%86%E5%BC%8F%E5%8C%96%E4%B8%BA%E9%83%A8%E5%88%86%E5%88%86%E5%BC%8F)
原式=a/(x+1)+b/(x+2)+c/(x+3)
=[a(x²+5x+6)+b(x²+4x+3)+c(x²+3x+2)]/(x+1)(x+2)(x+3)
=[(a+b+c)x²+(5a+4b+3c)x+(6a+3b+2c)]/(x+1)(x+2)(x+3)
所以
a+b+c=0
5a+4b+3c=1
6a+3b+2c=0
所以
a=-1/2.b=-2.c=5/2
原式=(-1/2)/(x+1)-2/(x+2)+(5/2)/(x+3)
再问: �� ����Ŷ ����1/(6(x 1)) 2/(15(x-2))-3/(10(x 3))
再答: Ŷ����������� a+b+c=0 5a+4b+3c=1 6a+3b+2c=0 ����a=-1/2��b=2��c=-3/2 ����Ĵ���ȻҲ����
=[a(x²+5x+6)+b(x²+4x+3)+c(x²+3x+2)]/(x+1)(x+2)(x+3)
=[(a+b+c)x²+(5a+4b+3c)x+(6a+3b+2c)]/(x+1)(x+2)(x+3)
所以
a+b+c=0
5a+4b+3c=1
6a+3b+2c=0
所以
a=-1/2.b=-2.c=5/2
原式=(-1/2)/(x+1)-2/(x+2)+(5/2)/(x+3)
再问: �� ����Ŷ ����1/(6(x 1)) 2/(15(x-2))-3/(10(x 3))
再答: Ŷ����������� a+b+c=0 5a+4b+3c=1 6a+3b+2c=0 ����a=-1/2��b=2��c=-3/2 ����Ĵ���ȻҲ����