2021 AMC 12A Spring 第 2 题

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2.

对实数 aabb 在什么条件下 a2+b2=a+b\sqrt{a^2 + b^2} = a + b 成立?

Under what conditions is a2+b2=a+b\sqrt{a^2 + b^2} = a + b true, where aa and bb are real numbers?

它永远不成立。

It is never true.

它成立当且仅当 ab=0ab = 0

It is true if and only if ab=0.ab = 0.

它成立当且仅当 a+b0a + b \ge 0

It is true if and only if a+b0.a + b \ge 0.

它成立当且仅当 ab=0ab = 0a+b0a + b \ge 0

It is true if and only if ab=0ab = 0 and a+b0.a + b \ge 0.

它总是成立。

It is always true.

答案:D
知识点:根式代数变形
难度评级:1200
解答:

因为 a2+b2\sqrt{a^2+b^2} 从不为负,等式成立必须有 a+b0a + b \ge 0。 两边平方得 a2+b2=(a+b)2a^2 + b^2 = (a+b)^2 =a2+2ab+b2= a^2 + 2ab + b^2, 化简为 2ab=02ab = 0, 即 ab=0ab = 0

反过来,如果 ab=0ab = 0,则 a2+b2=(a+b)2a^2 + b^2 = (a+b)^2;如果还满足 a+b0a + b \ge 0,那么 a2+b2=a+b=a+b\sqrt{a^2+b^2} = |a+b| = a+b。所以这两个条件合在一起正好是所需条件。

因此,正确答案是 D

Because a2+b2\sqrt{a^2+b^2} is never negative, equality requires a+b0.a + b \ge 0. Squaring both sides gives a2+b2=(a+b)2a^2 + b^2 = (a+b)^2 =a2+2ab+b2,= a^2 + 2ab + b^2, which simplifies to 2ab=0,2ab = 0, i.e. ab=0.ab = 0.

Conversely, if ab=0ab = 0 then a2+b2=(a+b)2,a^2 + b^2 = (a+b)^2, and if additionally a+b0a + b \ge 0 then a2+b2=a+b=a+b.\sqrt{a^2+b^2} = |a+b| = a+b. So both conditions together are exactly what is needed.

Thus, the correct answer is D.

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