若x^2+2y^2-2xy+4y+4=0,求x^y的值
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若x^2+2y^2-2xy+4y+4=0,求x^y的值若x^2+2y^2-2xy+4y+4=0,求x^y的值若x^2+2y^2-2xy+4y+4=0,求x^y的值(x+y)^2+(y+2)^2=0;所
若x^2+2y^2-2xy+4y+4=0,求x^y的值
若x^2+2y^2-2xy+4y+4=0,求x^y的值
若x^2+2y^2-2xy+4y+4=0,求x^y的值
(x+y)^2+(y+2)^2=0;所以 y=-2,x=2,所求为1/4
(x-y)^2+(y+2)^2=0
所以x=y=-2
所以x^y=1/4
∵x²+2y²-2xy-4y+4=0
x²+y²-2xy+y²-4y+4=0
(x²+y²-2xy)+(y²-4y+4)=0
(x-y)²+(y-2)²=0
且(x-y)²≥0,(y-2)²≥0
∴x-y=0,y-2...
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∵x²+2y²-2xy-4y+4=0
x²+y²-2xy+y²-4y+4=0
(x²+y²-2xy)+(y²-4y+4)=0
(x-y)²+(y-2)²=0
且(x-y)²≥0,(y-2)²≥0
∴x-y=0,y-2=0
解得x=y=2
x^y=4
收起
太简单了
配方可得(x-y)^2+(y+2)^2=0
两个非负数之和为0,所以这两个数都为0
所以x=y y=-2
再带进去求。。。
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