已知2x^2-2x-3=0，求代数式[（x-1）^2/x^2-1]+[2x^2-x/x+1]
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2x²-2x-3=0
所以2x²=2x+3
原式=(x-1)²/(x+1)(x-1)+(2x²-x)/(x+1)
=(x-1)/(x+1)+(2x²-x)/(x+1)
=(x-1+2x²-x)/(x+1)
=(2x²-1)/(x+1)
=(2x+3-1)/(x+1)
=2(x+1)/(x+1)
=2
所以2x²=2x+3
原式=(x-1)²/(x+1)(x-1)+(2x²-x)/(x+1)
=(x-1)/(x+1)+(2x²-x)/(x+1)
=(x-1+2x²-x)/(x+1)
=(2x²-1)/(x+1)
=(2x+3-1)/(x+1)
=2(x+1)/(x+1)
=2
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