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代数式x2-2xy+3y2-2x-2y+3的值的取值范围是 ______.
人气:347 ℃ 时间:2020-05-11 18:51:47
解答
原式=x2-2(y+1)x+3y2-2y+3
=x2-2(y+1)x+y2+2y+1+2y2-4y+2
=x2-2(y+1)x+(y+1)2+2(y-1)2
=(x-y-1)2+2(y-1)2
∴原式大于或等于0.
故本题答案为:大于或等于0.
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