(x+1)^2+x-2=1这道方程怎么解啊!其实原题目是：x+1 1----- + ----- =1x-2 x+1

(x+1)^2+x-2=1这道方程怎么解啊!其实原题目是：x+1 1----- + ----- =1x-2 x+1
(x+1)^2+x-2=1这道方程怎么解啊!
其实原题目是：
x+1 1
----- + ----- =1
x-2 x+1

(x+1)^2+x-2=1这道方程怎么解啊!其实原题目是：x+1 1----- + ----- =1x-2 x+1
(x+1)/(x-2)+1/(x+1)=1
(x-2+3)/(x-2)+1/(x+1)=1
(x-2)/(x-2)+3/(x-2)+1/(x+1)=1
1+3/(x-2)+1/(x+1)=1
3/(x-2)+1/(x+1)=0
3(x+1)/(x+1)(x-2)+(x-2)/(x+1)(x-2)=0
[3(x+1)+(x-2)]/(x+1)(x-2)=0
[3x+3+x-2]/(x+1)(x-2)=0
(4x+1)/(x+1)(x-2)=0
4x+1=0
x=-1/4
经检验x=-1/4是方程的解

令x+1=t t^2+t-3=1

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x=1/4