数学问题...急..2道

1 式变形为 x^2-2xy+y^2=0 so x/y=1
2 式变形为 (a-b)(a+2b)=0 so a=b or a=-2b
so (2a-b)/(2a+b)= 1/3 or 5/3

（x^2+y^2)/(x^2-xy+y^2)=2,求x/y
x^2+y^2=2x^2-2xy+2y^2
x^2-2xy+y^2=0
(x-y)^2=0
x=y
x/y=1
(2)a^2+ab-2b^2=0,
(a+2b)(a-b)=0
a=b或a=-2b
当a=b时,(2a-b)/(2a+b)=1/3
当a=-2b时,(2a-b)/(2a+b)=5/3

式变形为 x^2-2xy+y^2=0 so x/y=1