(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y
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(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y(x+√x∧2-2010)(y+√y^2-201
(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y
(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y
(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y
由题意知：X2-2010≥0 并且Y2-2010≥0
解得X≥√2010或X≤-√2010,Y≥√2010或Y≤-√2010
X=√2010,Y=√2010,原式=(√2010+0)(√2010+0)-2010=0成立
X=-√2010,Y=-√2010,原式=(-√2010+0)(-√2010+0)-2010=0成立
所以,本题有两组X=√2010,Y=√2010,X=-√2010,Y=-√2010
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