AMC10 2022 B
AMC10 2022 B · Q5
AMC10 2022 B · Q5. It mainly tests Exponents & radicals, Algebra misc.
What is the value of \[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\]
计算\[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}\]的值？
(A)
\sqrt3
\sqrt3
(B)
2
2
(C)
\sqrt{15}
\sqrt{15}
(D)
4
4
(E)
\sqrt{105}
\sqrt{105}
Answer
Correct choice: (B)
正确答案：（B）
Solution
We apply the difference of squares to the denominator, and then regroup factors:
\begin{align*} \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}} &= \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}\cdot\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}}{\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\frac43\cdot\frac65\cdot\frac87}}{\sqrt{\frac23\cdot\frac45\cdot\frac67}} \\ &= \frac{\sqrt{4\cdot6\cdot8}}{\sqrt{2\cdot4\cdot6}} \\ &= \frac{\sqrt8}{\sqrt2} \\ &= \boxed{\textbf{(B)}\ 2}. \end{align*}
我们对分母应用平方差公式，然后重新分组因子：
\begin{align*} \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}} &= \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}\cdot\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}}{\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\frac43\cdot\frac65\cdot\frac87}}{\sqrt{\frac23\cdot\frac45\cdot\frac67}} \\ &= \frac{\sqrt{4\cdot6\cdot8}}{\sqrt{2\cdot4\cdot6}} \\ &= \frac{\sqrt8}{\sqrt2} \\ &= \boxed{\textbf{(B)}\ 2}. \end{align*}
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