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解下列方程组 x^2+2xy+y^2=25 (x-y)^2-3x+3y+2=0
x^2+2xy+y^2=25
(x-y)^2-3x+3y+2=0
人气:482 ℃ 时间:2020-05-24 08:53:26
解答
(x+y)^2=25,则有x+y=正负5(x-y)^2-3(x-y)+2=0,(x-y-2)(x-y-1)=0,so x-y=2 or x-y=1有四种可能x+y=5 x-y=2 x=7/2 y=3/2x+y=5 x-y=1 x=3 y=2x+y=-5 x-y=2 x=-3/2 y=-7/2x+y=-5 x-y=1 x=-2 y=-3
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