先化简,再求值:1/x+2-（x²-4x+4/x²-x）除以（x+1-3/x-1)
先化简,再求值:1/x+2-（x²-4x+4/x²-x）除以（x+1-3/x-1),其中x满足x²+2x-4=0...
先化简,再求值:1/x+2-（x²-4x+4/x²-x）除以（x+1-3/x-1),其中x满足x²+2x-4=0
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原式=1/(x+2) - (x²-4x+4)/(x²-x) ÷ [ x+1 -3/(x-1) ]
=1/(x+2) -(x-2)²/[x(x-1)] ÷[ (x+1)(x-1)/(x-1) -3/(x-1) ]
=1/(x+2) -(x-2)²/[x(x-1)] ÷(x²-1-3)/(x-1)
=1/(x+2) -(x-2)²/[x(x-1)] ÷ (x²-4)/(x-1)
=1/(x+2) -(x-2)²/[x(x-1)] × (x-1)/[(x+2)(x-2)]
=1/(x+2) -(x-2)/[x(x+2)]
=x/[x(x+2)] -(x-2)/[x(x+2)]
=2/[x(x+2)]
=2/(x²+2x)
∵x²+2x-4=0
∴x²+2x=4
原式=2/4=1/2
=1/(x+2) -(x-2)²/[x(x-1)] ÷[ (x+1)(x-1)/(x-1) -3/(x-1) ]
=1/(x+2) -(x-2)²/[x(x-1)] ÷(x²-1-3)/(x-1)
=1/(x+2) -(x-2)²/[x(x-1)] ÷ (x²-4)/(x-1)
=1/(x+2) -(x-2)²/[x(x-1)] × (x-1)/[(x+2)(x-2)]
=1/(x+2) -(x-2)/[x(x+2)]
=x/[x(x+2)] -(x-2)/[x(x+2)]
=2/[x(x+2)]
=2/(x²+2x)
∵x²+2x-4=0
∴x²+2x=4
原式=2/4=1/2
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