AMC10 2008 A

AMC10 2008 A · Q7

AMC10 2008 A · Q7. It mainly tests Exponents & radicals, Manipulating equations.

The fraction $\frac{(3^{2008})^2 - (3^{2006})^2}{(3^{2007})^2 - (3^{2005})^2}$ simplifies to which of the following?

分式$\frac{(3^{2008})^2 - (3^{2006})^2}{(3^{2007})^2 - (3^{2005})^2}$化简为以下哪一项？

(A) 1 1

(B) $\frac{9}{4}$ $\frac{9}{4}$

(D) $\frac{9}{2}$ $\frac{9}{2}$

(E) 9 9

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): First note that $$ \frac{(3^{2008})^2-(3^{2006})^2}{(3^{2007})^2-(3^{2005})^2} = \frac{9^{2008}-9^{2006}}{9^{2007}-9^{2005}}. $$ Factoring $9^{2005}$ from each of the terms on the right side produces $$ \frac{9^{2008}-9^{2006}}{9^{2007}-9^{2005}} = \frac{9^{2005}\cdot 9^3-9^{2005}\cdot 9^1}{9^{2005}\cdot 9^2-9^{2005}\cdot 1} = \frac{9^{2005}}{9^{2005}}\cdot\frac{9^3-9}{9^2-1} = 9\cdot\frac{9^2-1}{9^2-1} = 9. $$

答案（E）：首先注意到 $$ \frac{(3^{2008})^2-(3^{2006})^2}{(3^{2007})^2-(3^{2005})^2} = \frac{9^{2008}-9^{2006}}{9^{2007}-9^{2005}}。 $$ 从右边每一项中提取 $9^{2005}$ 得到 $$ \frac{9^{2008}-9^{2006}}{9^{2007}-9^{2005}} = \frac{9^{2005}\cdot 9^3-9^{2005}\cdot 9^1}{9^{2005}\cdot 9^2-9^{2005}\cdot 1} = \frac{9^{2005}}{9^{2005}}\cdot\frac{9^3-9}{9^2-1} = 9\cdot\frac{9^2-1}{9^2-1} = 9。 $$

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