化简：（1）x-1/x+2÷x²-2x+1/x²-4+1/x-1（2）x-y/x+3y÷x²

解 化简：
(1)原式=[(x-1)/(x+2)]÷[(x²-2x+1)/(x²-4)]+1/(x-1)
=[(x-1)/(x+2)]÷{(x-1)²/[(x+2)(x-2)]}+1/(x-1)
=[(x-1)/(x+2)]×[(x+2)(x-2)/(x-1)²]+1/(x-1)
=(x-2)/(x-1)+1/(x-1)
=(x-2+1)/(x-1)
=(x-1)/(x-1)
=1
(2)原式=[(x-y)/(x+3y)]÷[(x²-y²)/(x²+6xy+9y²)]-2y/(x+y)
=[(x-y)/(x+3y)]÷[(x+y)(x-y)/(x+3y)²]-2y/(x+y)
=[(x-y)/(x+3y)]×{(x+3y)²/[(x+y)(x-y)]}-2y/(x+y)
=(x+3y)/(x+y)-2y/(x+y)
=(x+3y-2y)/(x+y)
=(x+y)/(x+y)
=1
解分式方程：
(2-x)/(x-3)+1/(3-x)=1
(2-x)/(x-3)-1/(x-3)=1
(2-x-1)/(x-3)=1
(1-x)/(x-3)=1
x-3=1-x
2x=4
x=2
经检验x=2是原方程的根

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