∵x=1-√2,y=1+√2
∴x+y=2，x-y=-2√2
原式=2x/[(x+y)(x-y)]*(x-y)^2/(2x)
=(x-y)/(x+y)
=-2√2/2
=-√2
[1/(x+y)+1/(x-y)]÷[2x/(x²-2xy+y²)]
=[[(x-y)+(x+y)]/(x+y)(x-y)]/[2x/(x-y)²]
=[2x/(x+y)(x-y)]/[2x/(x-y)²]
=(2x)(x-y)²/[2x(x+y)(x-y)]
=(x-y)/(x+y)
=[(1-√2)-(1+√2)]/[(1-√2)+(1+√2)]
=-2√2/2
=-√2
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