(1/x)+(1/y)=5,则(2x-5xy+2y)/(x+2xy+y)=?

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(1/x)+(1/y)=5,则(2x-5xy+2y)/(x+2xy+y)=?
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(1/x)+(1/y)=5,则(2x-5xy+2y)/(x+2xy+y)=?
(1/x)+(1/y)=5,则(2x-5xy+2y)/(x+2xy+y)=?

(1/x)+(1/y)=5,则(2x-5xy+2y)/(x+2xy+y)=?
1/x+1/y=(x+y)/xy=5
x+y=5xy
所以原式=[2(x+y)-5xy]/[(x+y)+2xy]
=[2(5xy)-5xy]/[(5xy)+2xy]
=5xy/7xy
=5/7

(1/x)+(1/y)=5
(x+y)/xy=5
x+y=5xy
(2x-5xy+2y)/(x+2xy+y)
=(10xy-5xy)/(5xy+2xy)
=5/7

(1/x)+(1/y)=5
分子分母上下同时除以xy：
(2x-5xy+2y)/(x+2xy+y)
=(2*5-5)/(5+2)
=5/7


（x＋y）/xy=5
两侧同时乘以xy，然后移项得：
x-5xy+y=0........(1)
于是（2x-5xy+2y）/（x+2xy+y)
=[2(x-5xy+y)+5xy]/[(x-5xy+y)+7xy](将（1）代入）
=5xy/7xy
=5/7

(1/x)+(1/y)=5，两边同时乘以xy，得y+x=5xy，
(2x-5xy+2y)/(x+2xy+y)=[2(x+y）-5xy]/（x+y+2xy）
将y+x=5xy代入[2(x+y）-5xy]/（x+y+2xy），
得（10xy-5xy）/（5xy+2xy）
=5xy/7xy
=5/7

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