【解方程1x^+3x+2+1x^2+5x+6+1x^2+7x+12+1x^2+x=4/21的解是】
2人问答
问题描述：
解方程1x^+3x+2+1x^2+5x+6+1x^2+7x+12+1x^2+x=4/21的解是
罗迒哉回答：
1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)+1/(x^2+x)=4/21
1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/x(x+1)=4/21
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)=4/21
1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)=4/21
1/x-1/(x+4)=4/21
(x+4-x)/x(x+4)=4/21
4/x(x+4)=4/21
1/x(x+4)=1/21
x(x+4)=21
x^2+4x-21=0
(x+7)(x-3)=0
x=-7或x=3
经检验x=-7或x=3是方程的解
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