解方程:(x+4/x+3)+(x+8/x+7)=(x/x-1)+(x+12/x+11)

如题所述

(x+4/x+3)+(x+8/x+7)=(x/x-1)+(x+12/x+11)
(x+8)/(x+7)-(x+12)/(x+11)=x/(x-1)-(x+4)/(x+3)
1+1/(x+7)-1-1/(x+11)=1+1/(x-1)-1-1/(x+3)
1/(x+7)-1/(x+11)=1/(x-1)-1/(x+3)
(x+11-x-7)/[(x+7)(x+11)]=(x+3-x+1)/[(x-1)(x+3)]
4/[(x+7)(x+11)]=4/[(x-1)(x+3)]
∴(x+7)(x+11)=(x-1)(x+3)
x²+18x+77=x²+2x-3
16x=-80
x=-5
经检验,是原方程的根。
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第1个回答  2012-04-02
原式=(x+3+1/x+3)+(x+7+1/x+7)=(x-1+1/x-1)+(x+11+1/x+11)
=1/(x+3)+1/(x+7)=1/(x-1)+1/(x+11)
(2x+10)/(x+3)(x+7)=(2x+10)/(x-1)(x+11)
分母不相等
所以2x+10=0
x=-5

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