1奥数题x-2y/x+2y=3,求x-2y/3(x+2y)-3(x-2y)/x-2y的差

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(x-2y)/(x+2y)=3
取倒数
(x+2y)/(x-2y)=1/3
所以原式=(1/3)[(x-2y)/(x+2y)]-3[(x+2y)/(x-2y)]
=(1/3)×3-3×(1/3)
=0

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