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题目已知x,y满足x2-2xy+y2-3~(1/2)x-3~(1/2)y+12=0,则 xy的最小值是_.错解 1 由 x2-2xy+y2-3~(1/2)x-3~(1/2)y+12=
The problem is known as x, y is x2-2xy+y2-3~(1/2)x-3~(1/2)y+12=0, then the minimum value of xy is _. False solution 1 is from x2-2xy +y2-3~(1/2)x-3~(1/2)y+12=