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题 实数ｘ、ｙ满足ｘ2 -2ｘｙ + ｙ2 -3ｘ -3ｙ +1 2 =0 ,则ｘｙ的最小值是 ( )。(Ａ) 1 2 (Ｂ) 4 51 6 (Ｃ) 2 14 (Ｄ)不存在。解法 1 由ｘ2 -2ｘｙ + ｙ2 -3ｘ -3ｙ + 1 2 =0 ,得ｘｙ =ｘ2 + ｙ2 -3ｘ -3ｙ+ 1 22=(ｘ -32 ) 2 + ( ｙ -32 ) 2 + 2 122 ≥2 14 ,
The real number x, y satisfies x2 -2xy + y2 -3x -3y +1 2 =0, then the minimum value of xy is (). (A) 1 2 (B) 4 51 6 (C) 2 14 (D) does not exist. Solution 1 From x2 -2xy + y2 -3x -3y + 1 2 =0, get xy =x2 + y2 -3x -3y+ 1 22=(x -32 ) 2 + ( y -32 ) 2 + 2 122 ≥2 14 ,