已知：3x-y=0,求（1+ 2y²/x²-y²)÷(1+ 2y/x-y)的值。
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（1+ 2y²/x²-y²)÷(1+ 2y/x-y)=((x²-y²+2y²)/(x²-y²))÷((x-y+ 2y)/(x-y))
=((x²+y²)/(x²-y²))÷((x+y)/(x-y))
=(x²+y²)/(x+y)²=(x²+9x²)/(x+3x)²=10/16=5/8
=((x²+y²)/(x²-y²))÷((x+y)/(x-y))
=(x²+y²)/(x+y)²=(x²+9x²)/(x+3x)²=10/16=5/8
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