设m等于1/5n求 2n/m+2n+m/2n-m+4mn/4n²-m²
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m=n/5
2n/（m+2n)＋m/(2n-m)+4mn/(4n²-m²)
=2n/（m+2n)＋m/(2n-m)+4mn/[(2n+m)(2n-m)]
=[2n(2n-m)+m(2n+m)+4mn]/[(2n+m)(2n-m)]
=(4n²-2mn+2mn+m²+4mn)/[(2n+m)(2n-m)]
=(4n²+4mn+m²)/[(2n+m)(2n-m)]
=(2n+m)²/[(2n+m)(2n-m)]
=(2n+m)/(2n-m)
=(2n+n/5)/(2n-n/5)
=(11n/5)/(9n/5)
=11/9
2n/（m+2n)＋m/(2n-m)+4mn/(4n²-m²)
=2n/（m+2n)＋m/(2n-m)+4mn/[(2n+m)(2n-m)]
=[2n(2n-m)+m(2n+m)+4mn]/[(2n+m)(2n-m)]
=(4n²-2mn+2mn+m²+4mn)/[(2n+m)(2n-m)]
=(4n²+4mn+m²)/[(2n+m)(2n-m)]
=(2n+m)²/[(2n+m)(2n-m)]
=(2n+m)/(2n-m)
=(2n+n/5)/(2n-n/5)
=(11n/5)/(9n/5)
=11/9
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