Question

对实数 $x$ 和 $y$ ， $(xy-1)^2+(x+y)^2$ 的最小可能值是多少？

What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y?$

$0$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$1$

$2$

答案：D

知识点：最优化代数变形

难度评级：770

视频讲解：

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文字解答：

展开得每个平方项都非负，因此除 $1$ 外的各项都取 $0$ 时，总和取得最小值 $1$ 。 $\begin{gathered} x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 \\ = x^2y^2 + x^2 + y^2 + 1. \end{gathered}$

当 $x = y = 0$ 时可以取到这个值。

所以正确答案是 D。

Expanding, we get $\begin{gathered} x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 \\ = x^2y^2 + x^2 + y^2 + 1. \end{gathered}$ Note that every square must be non-negative. Therefore, the minimum value is when all the terms except $1$ are $0,$ making the sum $1.$

This is attainable when $x = y = 0.$

Thus, D is the correct answer.

2021 AMC 10A Spring 第 9 题

答案：D

难度评级：770

视频讲解：

文字解答：

其他年份的第 9 题