问题 解答题
x2-9y2=0
x2-2xy+y2=4
答案

原方程组变形为:

(x-3y)(x+3y)=0
(x-y-2)(x-y+2)=0

原方程组变为四个方程组为:

x-3y=0
x-y-2=0
x-3y=0
x-y+2=0
x+3y=0
x-y-2=0
x+3y=0
x-y+2=0

解这四个方程组为:

x1=3
y1=1
x2=-3
y2=-1
x3=
3
2
y3=-
1
2
x4=-
3
2
y4=
1
2

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