如题。谢谢了
过程了
解：由2x^2+4xy+4y^2+8x+12y+10=0得
x^2+2xy+2y^2+4x+6y+5=0
x^2+2(y+2)x+(y+2)^2-(y+2)^2+2y^2+6y+5=0
(x+y+2)^2+(y+1)^2=0
于是x+y+2=0 y+1=0
故x+y=-2
x^2+2xy+2y^2+4x+6y+5=0
x^2+2(y+2)x+(y+2)^2-(y+2)^2+2y^2+6y+5=0
(x+y+2)^2+(y+1)^2=0
于是x+y+2=0 y+1=0
故x+y=-2
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第1个回答 2019-01-14
2X^2+4XY+4Y^2+8X+12Y+10
=2(x+y)^2+2Y^2+8x+12y+10
=2(x+y)^2+8(x+y)+8+[2Y^2+4y+2]
=2[(x+y)+2]^2+2(y+1)^2=0
2[(x+y)+2]^2>=0
2(y+1)^2>=0
所以(x+y)+2＝0，且y+1＝0
所以x+y＝-2
=2(x+y)^2+2Y^2+8x+12y+10
=2(x+y)^2+8(x+y)+8+[2Y^2+4y+2]
=2[(x+y)+2]^2+2(y+1)^2=0
2[(x+y)+2]^2>=0
2(y+1)^2>=0
所以(x+y)+2＝0，且y+1＝0
所以x+y＝-2
