先化简,再求值:1-[(x)-1/1-x]2除以x2-x+1/x2-2x+1,其中x2-x+5=0
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解：
原式=1-[(x)-1/(1-x)]²÷[(x²-x+1)/(x²-2x+1)]
=1-[x(1-x)/(1-x)-1/(1-x)]²÷[(x²-x+1)/(x-1)²]
=1-[(x-x²-1)/(1-x)]²×[(x-1)²/(x²-x+1)]
=1-[(x²-x+1)/(x-1)]²×[(x-1)²/(x²-x+1)]
=1-[(x²-x+1)²/(x-1)²]×[(x-1)²/(x²-x+1)]
=1-(x²-x+1)
=-(x²-x)
∵ x²-x+5=0
∴ x²-x=-5
原式=-(-5)
=5
原式=1-[(x)-1/(1-x)]²÷[(x²-x+1)/(x²-2x+1)]
=1-[x(1-x)/(1-x)-1/(1-x)]²÷[(x²-x+1)/(x-1)²]
=1-[(x-x²-1)/(1-x)]²×[(x-1)²/(x²-x+1)]
=1-[(x²-x+1)/(x-1)]²×[(x-1)²/(x²-x+1)]
=1-[(x²-x+1)²/(x-1)²]×[(x-1)²/(x²-x+1)]
=1-(x²-x+1)
=-(x²-x)
∵ x²-x+5=0
∴ x²-x=-5
原式=-(-5)
=5
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