AMC10 2018 A

AMC10 2018 A · Q10

AMC10 2018 A · Q10. It mainly tests Exponents & radicals, Manipulating equations.

Suppose that real number $x$ satisfies $\sqrt{49 - x^2} - \sqrt{25 - x^2} = 3$. What is the value of $\sqrt{49 - x^2} + \sqrt{25 - x^2}$?

设实数 $x$ 满足 $\sqrt{49 - x^2} - \sqrt{25 - x^2} = 3$。求 $\sqrt{49 - x^2} + \sqrt{25 - x^2}$ 的值。

(A) 8 8

(B) $\sqrt{33} + 3$ $\sqrt{33} + 3$

(D) $2\sqrt{10} + 4$ $2\sqrt{10} + 4$

(E) 12 12

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let $a=\sqrt{49-x^2}-\sqrt{25-x^2}$ and $b=\sqrt{49-x^2}+\sqrt{25-x^2}$. Then $ab=(49-x^2)-(25-x^2)=24$, so $b=\dfrac{24}{a}=\dfrac{24}{3}=8$.

答案（A）：设 $a=\sqrt{49-x^2}-\sqrt{25-x^2}$，$b=\sqrt{49-x^2}+\sqrt{25-x^2}$。则 $ab=(49-x^2)-(25-x^2)=24$，所以 $b=\dfrac{24}{a}=\dfrac{24}{3}=8$。

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