[3-2x/1-x]-1=[x+2/x-3]怎么做

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[3-2x/1-x]-1=[x+2/x-3]怎么做
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[3-2x/1-x]-1=[x+2/x-3]怎么做
[3-2x/1-x]-1=[x+2/x-3]怎么做

[3-2x/1-x]-1=[x+2/x-3]怎么做
这个式子有问题啊
应该是（3-2x）/(1-x)-1=(x+2)/(x-3)
先通分[(3-2x)-(1-x)]/(1-x)=(x+2)/(x-3)
化简（2-x)/(1-x)=(x+2)/(x-3)
即(2-x)*(x-3)=(x+2)*(1-x)
x=4/3

先通分,再分类讨论,分母不为零

(3-2x)/(1-x)-(1-x)/(1-x)=(x+2)/(x-3)
得(2-x)/(1-x)=(x+2)/(x-3)
得（2-x)(x-3)=(1-x)(x+2)
得5x-6=2-x
得6x=8
得x=4/3

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