Question

数学分式计算

若x^2+xy-2y^2=0,则(x^2+3xy+3y^2)/(x^2+y^2)=( )
请说明一下过程。O(∩_∩)O谢谢

Answer 1

x^2+xy-2y^2=0
(x+2y)(x-y)=0
x=-2y或x=y

x=-2y
原式=(4y^2-6y^2+3y^2)/(4y^2+y^2)=y^2/5y^2=1/5

x=y
原式=(y^2+3y^2+3y^2)/(y^2+y^2)=7y^2/2y^2=7/2

Answer 2

x^2+xy-2y^2=0,
(x+2y)(x-y)=0
x=-2y或x=y

当x=-2y时
(x^2+3xy+3y^2)/(x^2+y^2)
=(4y^2-6y^2+3y^2)/(4y^2+y^2)
=y^2/5y^2
=1/5

当x=y时
(x^2+3xy+3y^2)/(x^2+y^2)
=(y^2+3y^2+3y^2)/(y^2+y^2)
=7y^2/2y^2
=7/2