数学先化简再求值！
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原式=y(x+2y)(x-2y)/(x+2y)²*(4xy+x²-2xy)/(x-2y)
=y(x-2y)/(x+2y)*(x²+2xy)/(x-2y)
=y(x-2y)/(x+2y)*x(x+2y)/(x-2y)
=xy
=(√2-1)(√2+1)
=2-1
=1
=y(x-2y)/(x+2y)*(x²+2xy)/(x-2y)
=y(x-2y)/(x+2y)*x(x+2y)/(x-2y)
=xy
=(√2-1)(√2+1)
=2-1
=1
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