已知x²+2x-1=0,试求代数式1/(x+1)-(x+3)/(x²-1)•(x²-2x+1)/(x²+4x+3)

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已知x²+2x-1=0,试求代数式1/(x+1)-(x+3)/(x²-1)•(x²-2x+1)/(x²+4x+3)
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已知x²+2x-1=0,试求代数式1/(x+1)-(x+3)/(x²-1)•(x²-2x+1)/(x²+4x+3)
已知x²+2x-1=0,试求代数式1/(x+1)-(x+3)/(x²-1)•(x²-2x+1)/(x²+4x+3)

已知x²+2x-1=0,试求代数式1/(x+1)-(x+3)/(x²-1)•(x²-2x+1)/(x²+4x+3)
1/(x+1)-(x+3)/(x²-1)•(x²-2x+1)/(x²+4x+3)
=1/(x+1)-(x+3)/[(x-1)(x+1)] * (x+1)^2/[(x+1)(x+3)]
=1/(x+1)-1/(x-1)
=2/(1-x^2)
又x²+2x-1=0得
2x=1-x^2
原式=2/2x=1/x
x=-1-√2或-1+√2

楼上的错了吧。。。