已知x=2-根号3,求（x-1分之1-2x+x平方）-（x平方-x 分之根号x平方-2x+1）的值.
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x=2-√3
x-1=1-√3﹤0
原式=(1-2x+x²)/(x-1)-√(x²-2x+1)/(x²-x)
=(x-1)²/(x-1)-√(x-1)²/[x(x-1)]
=x-1-|x-1|/[x(x-1)]
=x-1+(x-1)/[x(x-1)]
=x-1+(1/x)
=1-√3+1/(2-√3)
=1-√3+(2+√3)/[(2+√3)(2-√3)]
=1-3+(2+√3)/(2²-√3²)
=1-√3+(2+√3)/(4-3)
=1-√3+2+√3
=3
x-1=1-√3﹤0
原式=(1-2x+x²)/(x-1)-√(x²-2x+1)/(x²-x)
=(x-1)²/(x-1)-√(x-1)²/[x(x-1)]
=x-1-|x-1|/[x(x-1)]
=x-1+(x-1)/[x(x-1)]
=x-1+(1/x)
=1-√3+1/(2-√3)
=1-√3+(2+√3)/[(2+√3)(2-√3)]
=1-3+(2+√3)/(2²-√3²)
=1-√3+(2+√3)/(4-3)
=1-√3+2+√3
=3
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