已知x/y=2,求（x+y）2-4xy/x2+y2

已知x/y=2,求（x+y）2-4xy/x2+y2
已知x/y=2,求（x+y）2-4xy/x2+y2

已知x/y=2,求（x+y）2-4xy/x2+y2
若题目是[（x+y）^2-4xy]/(x^2+y^2),则
分子分母同时除以y^2,化成：[(x/y+1)^2-4x/y]/[(x/y)^2+1}
因为：x/y=2,代入上式可得
原式等=[（2+1）^2-4*2]/(2^2+1)
=(9-8)/(4+1)
=1/5
=0.2

已知x/y=2,求((x+y)^2-4xy)/(x^2+y^2)
((x+y)^2-4xy)/(x^2+y^2)
=(x^2+2xy+y^2-4xy)/(x^2+y^2)
=(x^2+y^2-2xy)/(x^2+y^2)
=1-2xy/(x^2+y^2)
=1-2/(x^2/xy+y^2/xy)
=1-2/(x/y+y/x)
=1-2/(2+1/2)
=1-4/5
=1/5