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已知X.Y是实数且满足X^2+XY+Y^2-2=0,设M=X^2-XY+Y^2,则M的取值范围是
人气:129 ℃ 时间:2019-10-17 03:54:15
解答
X^2+XY+Y^2-2=0X^2+XY+Y^2=2x^2+y^2=2-xy又:(x-y)^2≥0xy≤(x^2+y^2)/2=(2-xy)/2=1-xy/2(1+1/2)xy≤1xy≤2/3∴M=X^2-XY+Y^2=(X^2+XY+Y^2)-2xy=2-2xy ≥ 2-2*2/3=2/3即M≥2/3
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