一个数的倒数通过交换分子和分母得到。例如， \( \frac{2}{3} \) 的倒数是 \( \frac{3}{2} \) ，而 2(注意 2 可以写成 \( \frac{2}{1} \) )的倒数是 \( \frac{1}{2} \) 。(8)
分数之和
\[ \frac{1}{n\left( {n + 1}\right) } = \frac{1}{n} - \frac{1}{n + 1} \tag{9} \]
\( \frac{1}{n\left( {n + k}\right) } = \frac{1}{k}\left( {\frac{1}{n} - \frac{1}{n + k}}\right) \) (10)
连分数
一个数的简单连分数表示为:
\[ {a}_{0} + \frac{1}{{a}_{1} + \frac{1}{{a}_{2} + \frac{1}{{a}_{3} + \frac{1}{{a}_{4} + \ldots }}}} \]
其中 \( {a}_{0} \) 为整数，其余 \( {a}_{i} \) 均为正整数。
3. 指数与根式
指数法则
幂法则:
\[ {a}^{m} \times {a}^{n} = {a}^{m + n} \tag{11} \]
\[ {\left( {a}^{m}\right) }^{n} = {a}^{mn} \tag{12} \]
\[ {\left( ab\right) }^{n} = {a}^{n}{b}^{n} \tag{13} \]
\[ \frac{{a}^{m}}{{b}^{m}} = {\left( \frac{a}{b}\right) }^{m} \tag{14} \]
商法则:
\[ \frac{{a}^{m}}{{a}^{n}} = {a}^{m - n} \tag{15} \]
根式法则
\[ \sqrt[n]{{a}^{m}} = {\left( \sqrt[n]{a}\right) }^{m} = {a}^{\frac{m}{n}} \tag{16} \]
\( \sqrt[n]{\sqrt[m]{a}} = {\left( {a}^{\frac{1}{m}}\right) }^{\frac{1}{n}} = {a}^{\frac{1}{n} \times \frac{1}{m}} = {a}^{\frac{1}{nm}} = \sqrt[{nm}]{a} \) (17)
EXAMPLES
例 1. 若 \( m + \frac{1}{m} = 4 \) ，求 \( {m}^{2} + \frac{1}{{m}^{2}} \) 。
(A) 4 (B) 8 (C) 14 (D) 12 (E) 10
解:(C)。
方法一:
由于 \( {\left( x + y\right) }^{2} = {x}^{2} + {2xy} + {y}^{2} \) ，我们有:
\( {x}^{2} + {y}^{2} = {\left( x + y\right) }^{2} - {2xy} \) (1)
在 (1) 中令 \( x = m \) 和 \( y = \frac{1}{m} \) : \( {m}^{2} + \frac{1}{{m}^{2}} = {\left( m + \frac{1}{m}\right) }^{2} - {2m} \times \frac{1}{m} = {4}^{2} - 2 = {14} \) 。
方法二:
\[ {m}^{2} + \frac{1}{{m}^{2}} = {m}^{2} + 2 - 2 + \frac{1}{{m}^{2}} = {m}^{2} + 2 \times m \times \frac{1}{m} - 2 + \frac{1}{{m}^{2}} \]
\[ = \left\lbrack {{m}^{2} + 2 \times m \times \frac{1}{m} + {\left( \frac{1}{m}\right) }^{2}}\right\rbrack - 2 = {\left( m + \frac{1}{m}\right) }^{2} - 2 = {4}^{2} - 2 = {14}. \]
例2. 若 \( m + \frac{1}{m} = 2 \) ，求 \( {m}^{3} + \frac{1}{{m}^{3}} \) 。
(A) 4 (B) 8 (C) 14 (D) 2 (E) 10
解答:(D)。
方法一:
将 \( m + \frac{1}{m} = 2 \) 两边同乘 \( m \) ，得
\( {m}^{2} + 1 = {2m} \Rightarrow \;{m}^{2} - {2m} + 1 = 0\; \Rightarrow \;{\left( m - 1\right) }^{2} = 0. \)
解此方程，得 \( m = 1 \) 。
因此 \( {m}^{3} + \frac{1}{{m}^{3}} = {1}^{3} + \frac{1}{{1}^{3}} = 1 + 1 = 2 \) 。
方法二:
我们知道 \( {\left( x + y\right) }^{3} = {x}^{3} + 3{x}^{2}y + {3x}{y}^{2} + {y}^{3} = {x}^{3} + {y}^{3} + {3xy}\left( {x + y}\right) \) 。
所以 \( {x}^{3} + {y}^{3} = {\left( x + y\right) }^{3} - {3xy}\left( {x + y}\right) \) (1)
将 \( x = m \) 和 \( y = \frac{1}{m} \) 代入(1)得:
\( {m}^{3} + \frac{1}{{m}^{3}} = {\left( m + \frac{1}{m}\right) }^{3} - {3m} \times \frac{1}{m}\left( {m + \frac{1}{m}}\right) = {2}^{3} - 3 \times 2 = 2. \)
方法三:
我们知道 \( {x}^{3} + {y}^{3} = \left( {x + y}\right) \left( {{x}^{2} - {xy} + {y}^{2}}\right) \) 。
将 \( x = m \) 和 \( y = \frac{1}{m} \) 代入上式，得:
\[ {m}^{3} + \frac{1}{{m}^{3}} = \left( {m + \frac{1}{m}}\right) \left( {{m}^{2} - m \cdot \frac{1}{m} + \frac{1}{{m}^{2}}}\right) = 2\left\lbrack {{\left( m + \frac{1}{m}\right) }^{2} - 3}\right\rbrack = 2\left( {{2}^{2} - 3}\right) = 2 \]
方法四:
将 \( m + \frac{1}{m} = 2 \) 两边同乘以 \( \left( {{m}^{2} + \frac{1}{{m}^{2}} - 1}\right) \) ，得到
\[ \left( {{m}^{2} + \frac{1}{{m}^{2}} - 1}\right) \left( {m + \frac{1}{m}}\right) = 2\left( {{m}^{2} + \frac{1}{{m}^{2}} - 1}\right) \; \Rightarrow \;{m}^{3} + \frac{1}{{m}^{3}} = \]
\[ 2\left\lbrack {{\left( m + \frac{1}{m}\right) }^{2} - 2 - 1}\right\rbrack = 2. \]
例3. 若 \( x = \frac{3 + \sqrt{13}}{2} \) ，求 \( x - \frac{1}{x} \) 的值。
(A) 3 (B) 4 (C) 10 (D) 12 (E) 1
解:(A)。
方法一:
由于 \( x = \frac{3 + \sqrt{13}}{2} \Rightarrow \;{2x} - 3 = \sqrt{13} \)
\( \Rightarrow \;4{x}^{2} - {12x} - 4 = 0\; \Rightarrow \;{x}^{2} - {3x} - 1 = 0 \) (1)
将(1)式两边同除以 \( x : \Rightarrow x - 3 - \frac{1}{x} = 0\; \Rightarrow \;x - \frac{1}{x} = 3 \) 。
方法二:
\[ x = \frac{3 + \sqrt{13}}{2} \Rightarrow \frac{1}{x} = \frac{2}{3 + \sqrt{13}} = \frac{2}{\sqrt{13} + 3} = \frac{2\left( {\sqrt{13} - 3}\right) }{\left( {\sqrt{13} + 3}\right) \left( {\sqrt{13} - 3}\right) } \]
\[ = \frac{2\left( {\sqrt{13} - 3}\right) }{{13} - 9} = \frac{2\left( {\sqrt{13} - 3}\right) }{4} = \frac{\sqrt{13} - 3}{2}. \]
\[ x - \frac{1}{x} = \frac{3 + \sqrt{13}}{2} - \frac{\sqrt{13} - 3}{2} = = \frac{3 + 3}{2} = 3. \]
例4. 化简下列有理函数: \( \frac{{x}^{3} - {y}^{3}}{{x}^{4} + {x}^{2}{y}^{2} + {y}^{4}} \) 。
(A) \( \frac{x - y}{{x}^{2} + {xy} + {y}^{2}} \) (B) \( \frac{x - y}{{x}^{2} - {xy} + {y}^{2}} \) (C) \( \frac{1}{{x}^{2} - {xy} + {y}^{2}} \) (D) \( \frac{1}{{x}^{2} + {y}^{2}} \) (E) none of
these
解:(B)。
\[ \frac{{x}^{3} - {y}^{3}}{{x}^{4} + {x}^{2}{y}^{2} + {y}^{4}} = \frac{\left( {x - y}\right) \left( {{x}^{2} + {xy} + {y}^{2}}\right) }{{x}^{4} + {x}^{2}{y}^{2} + {y}^{4}} = \frac{\left( {x - y}\right) \left( {{x}^{2} + {xy} + {y}^{2}}\right) }{\left( {{x}^{2} + {xy} + {y}^{2}}\right) \left( {{x}^{2} - {xy} + {y}^{2}}\right) } = \]
\[ \frac{\left( x - y\right) }{\left( {x}^{2} - xy + {y}^{2}\right) }\text{.} \]
例5. 下列哪一项与
\( \frac{2 - 6 + {10} - {14} + {18} - {22} + {26}}{3 - 9 + {15} - {21} + {27} - {33} + {39}}? \)
(A) -1 (B) \( - \frac{2}{3} \) (C) \( \frac{2}{3} \) (D) 1 (E) \( \frac{14}{3} \)
解:(C)。
\[ \frac{2 - 6 + {10} - {14} + {18} - {22} + {26}}{3 - 9 + {15} - {21} + {27} - {33} + {39}} = \frac{2\left( {1 - 3 + 5 - 7 + 9 - {11} + {13}}\right) }{3\left( {1 - 3 + 5 - 7 + 9 - {11} + {13}}\right) } = \frac{2}{3}. \]
例6. \( \frac{1 + 3 + 5}{2 + 4 + 6} - \frac{2 + 4 + 6}{1 + 3 + 5} \) 是多少？
(A) 1 (B) \( - \frac{5}{36} \) (C) \( - \frac{7}{12} \) (D) \( \frac{7}{12} \) (E) \( \frac{34}{3} \)
解答:(C)。
\[ \frac{1 + 3 + 5}{2 + 4 + 6} - \frac{2 + 4 + 6}{1 + 3 + 5} = \frac{9}{12} - \frac{12}{9} = \frac{3}{4} - \frac{4}{3} = \frac{9 - {16}}{12} = - \frac{7}{12}. \]
例7. 下列哪一项等于 \( \frac{\frac{1}{2} - \frac{1}{3}}{\frac{1}{3} - \frac{1}{4}} \) ？
(A) \( \frac{1}{3} \) (B) \( \frac{1}{4} \) (C) 2 (D) \( \frac{3}{2} \) (E) \( \frac{4}{3} \)
解答:(C)。
\( \frac{\frac{1}{2} - \frac{1}{3}}{\frac{1}{3} - \frac{1}{4}} = \frac{\frac{3 - 2}{4 - 3}}{\frac{4 - 3}{12}} = \frac{\frac{1}{6}}{\frac{1}{12}} = \frac{12}{6} = 2 \)
例8. (2009 AMC 10 A) 下列哪一项等于 \( 1 + \frac{1}{1 + \frac{1}{1 + 1}} \) ？
(A) \( \frac{5}{4} \) (B) \( \frac{3}{2} \) (C) \( \frac{5}{3} \) (D) 2 (E) 3
解答:(C)。
\( 1 + \frac{1}{1 + \frac{1}{1 + 1}} = 1 + \frac{1}{1 + \frac{1}{2}} = 1 + \frac{1}{\frac{3}{2}} = 1 + \frac{2}{3} = \frac{5}{3}. \)
例9. 下列哪一项等于该乘积
\( \frac{6}{3} \times \frac{9}{6} \times \frac{12}{9} \times \cdots \times \frac{{3n} + 3}{3n}\cdots \times \frac{2019}{2016}? \)
(A) 673 (B) 1346 (C) 2019 (D) 2016 (E) 4038
解答:(A)。
\[ \frac{6}{3} \times \frac{9}{6} \times \frac{12}{9} \times \cdots \times \frac{{3n} + 3}{3n}\cdots \times \frac{2019}{2016} = \frac{6}{3} \times \frac{9}{6} \times \frac{12}{9} \times \cdots \times \frac{{3n} + 3}{3n}\cdots \times \frac{2019}{2016} = \]
\[ \frac{2019}{3} = {673} \]
例10. (2002 AMC 10) 对于非零数 \( a, b \) 和 \( c \) ，定义
\( \left( {a, b, c}\right) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \) 。求(2,12,9)的值？
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
解答:(C)。
\( \left( {a, b, c}\right) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \Rightarrow \) \( + \frac{c}{a} \Rightarrow \;\left( {2,{12},9}\right) = \frac{2}{12} + \frac{12}{9} + \frac{9}{2} = \frac{56}{12} + \frac{12}{9} = \frac{14}{3} + \frac{4}{3} = \frac{18}{3} = 6 \)
例11. 对于正整数 \( a, b \) 和 \( c \) ，若 \( \left( {a, b, c}\right) = \frac{abc}{a + b + c} \) ，求(3,5,7)。
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
解答:(D)。
\[ \left( {a, b, c}\right) = \frac{abc}{a + b + c}\; \Rightarrow \;\left( {3,5,7}\right) = \frac{3 \times 5 \times 7}{3 + 5 + 7} = \frac{105}{15} = 7. \]
例12. 求和: \( \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{{49} \times {50}} \) 。
(A) \( \frac{5}{12} \) (B) \( \frac{49}{50} \) (C) \( \frac{5}{13} \) (D) \( \frac{50}{49} \) (E) 1
解答:(B)。
\[ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{{49} \times {50}} = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{50} = 1 - \frac{1}{50} = \frac{49}{50} \]
例13. 求和: \( \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \cdots + \frac{1}{{11} \times {13}} \) 。
(A) \( \frac{6}{13} \) (B) \( \frac{3}{2} \) (C) \( \frac{5}{13} \) (D) \( \frac{5}{4} \) (E) 3
解答:(A)。
\[ \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \cdots + \frac{1}{{11} \times {13}} = \frac{1}{2}\left( {\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \ldots + \frac{1}{11} - \frac{1}{13}}\right) = \frac{1}{2} \times \left( {1 - \frac{1}{13}}\right) = \frac{6}{13}. \]
例14. 计算: \( \frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \cdots + \frac{1}{{98} \times {100}} \) 。
(A) \( \frac{49}{50} \) (B) \( \frac{3}{2} \) (C) \( \frac{5}{13} \) (D) \( \frac{49}{200} \) (E) \( \frac{2}{3} \)
解答:(D)。
方法一:
\[ \frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \cdots + \frac{1}{{98} \times {100}} \]
\[ = \frac{1}{2}\left( {\frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{6} + \frac{1}{6} - \ldots + \frac{1}{98} - \frac{1}{100}}\right) = \frac{1}{2} \times \left( {\frac{1}{2} - \frac{1}{100}}\right) = \frac{1}{2}\left( {\frac{50}{100} - \frac{1}{100}}\right) = \frac{49}{200}. \]
方法二:
\[ \frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \cdots + \frac{1}{{98} \times {100}} = \frac{1}{4}\left( {\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{{49} \times {50}}}\right) \]
\[ = \frac{1}{4}\left( {\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \cdots + \frac{1}{49} - \frac{1}{50}}\right) = \frac{1}{4}\left( {1 - \frac{1}{50}}\right) = \frac{49}{200}\text{.} \]
例15. 计算: \( \frac{3}{1 \times 4} + \frac{3}{4 \times 7} + \frac{3}{7 \times {10}} + \cdots + \frac{3}{{19} \times {22}} \) 。
(A) \( \frac{1}{2} \) (B) \( \frac{21}{22} \) (C) \( \frac{5}{13} \) (D) \( \frac{19}{22} \) (E) \( \frac{3}{22} \)
解答:(B)
方法一:
\[ \frac{3}{1 \times 4} + \frac{3}{4 \times 7} + \frac{3}{7 \times {10}} + \cdots + \frac{3}{{19} \times {22}} = \]
\[ \frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7}\ldots - \frac{1}{19} + \frac{1}{19} - \frac{1}{22} = 1 - \frac{1}{22} = \frac{21}{22} \]
方法二:
\[ \frac{3}{1 \times 4} + \frac{3}{4 \times 7} + \frac{3}{7 \times {10}} + \cdots + \frac{3}{{19} \times {22}} = 3\left( {\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times {10}} + \cdots + \frac{1}{{19} \times {22}}}\right) \]
\[ = 3\left\lbrack \left( {\frac{1}{3}\left( {\frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7}\ldots - \frac{1}{19} + \frac{1}{19} - \frac{1}{22}}\right) }\right) \right\rbrack = 1 - \frac{1}{22} = \frac{21}{22}\text{.} \]
例16. 计算 \( \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - \cdots }}}}} \)
(A) -1 (B) \( - 1/2 \) (C) 0 (D) \( 1/2 \) (E) 1
解答:(E)。
\[ \text{Let}\frac{1}{2 - \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - \frac{1}{2 - \cdots }}}}} = x.\;\frac{1}{2 - x} = x\; \Rightarrow \;{x}^{2} - {2x} + 1 = 0 \Rightarrow \]
\[ {\left( x - 1\right) }^{2} = 0\; \Rightarrow \;x = 1. \]
例17. 若 \( \frac{1}{x} - \frac{1}{y} = 3 \) ，求 \( \frac{{2x} + {3xy} - {2y}}{x - {2xy} - y} \) 。(A) \( \frac{1}{2} \) (B) \( \frac{3}{4} \) (C) \( \frac{5}{13} \) (D) \( \frac{19}{22} \) (E) \( \frac{3}{5} \)
解答:(E)。
方法一:
由于分母不能为0，可知 \( x \neq 0, y \neq 0 \) ，且 \( x \)
\( - {2xy} - y \neq 0 \) .
将 \( \frac{{2x} + {3xy} - {2y}}{x - {2xy} - y} \) 的每一项除以 \( {xy} \) ，得到:
\( \frac{\frac{2}{y} - \frac{2}{x} + 3}{\frac{1}{y} - \frac{1}{x} - 2} = \frac{-2\left( {\frac{1}{x} - \frac{1}{y}}\right) + 3}{-\left( {\frac{1}{x} - \frac{1}{y}}\right) - 2} = \frac{-2 \times 3 + 3}{-3 - 2} = \frac{3}{5}. \)
方法二:
将 \( \frac{1}{x} - \frac{1}{y} = 3 \) 两边同乘以 \( {xy} \) ，得到 \( y - x = {3xy} \) 。
因此 \( \frac{{2x} + {3xy} - {2y}}{x - {2xy} - y} = \frac{-2\left( {y - x}\right) + {3xy}}{-\left( {y - x}\right) - {2xy}} = \frac{-{6xy} + {3xy}}{-{3xy} - {2xy}} = \frac{3}{5} \) 。
方法三:
在方程 \( \frac{1}{x} - \frac{1}{y} = 3 \) 中解出 \( x \) ，得到 \( x = \frac{y}{{3y} + 1} \) 。
因此
\[ \frac{{2x} + {3xy} - {2y}}{x - {2xy} - y} = \frac{\frac{2y}{{3y} + 1} + \frac{3{y}^{2}}{{3y} + 1} - \frac{6{y}^{2} + {2y}}{{3y} + 1}}{\frac{y}{{3y} + 1} - \frac{2{y}^{2}}{{3y} + 1} - \frac{3{y}^{2} + y}{{3y} + 1}} = \frac{{2y} + 3{y}^{2} - 6{y}^{2} - {2y}}{y - 2{y}^{2} - 3{y}^{2} - y} = \frac{-3{y}^{2}}{-5{y}^{2}} = \frac{3}{5}. \]
方法4:
设 \( \frac{{2x} + {3xy} - {2y}}{x - {2xy} - y} = k \) 。
\( {2x} + {3xy} - {2y} = k\left( {x - {2xy} - y}\right) \)
\( \Rightarrow \;{2x} + {3xy} - {2y} = {kx} - {2kxy} - {ky} \)
\[ \Rightarrow \;{2y} - {2x} - {ky} + {kx} = {3xy} + {2kxy} \]
\( \Rightarrow \;\left( {2 - k}\right) \left( {y - x}\right) = \left( {3 + {2k}}\right) {xy} \) (1)
由于 \( \frac{1}{x} - \frac{1}{y} = 3 \) ，两边同乘 \( {xy} \) 得 \( y - x = {3xy} \) 。(2)
将方程(2)除以方程(1)得
(1) \( \div \left( 2\right) : 2 - k = \frac{3 + {2k}}{3}\; \Rightarrow \;k = \frac{3}{5} \) .
\( k = \frac{{2x} + {3xy} - {2y}}{{x}^{2} - {2xy} - {y}^{2}} = \frac{3}{5}. \)
例18. 数 \( {40000} \times {40000}^{40000} \) 与下列哪一项相同？
(A) \( {200}^{80002} \) (B) \( {80000}^{40000} \) (C) \( {200}^{40001} \)
(D) \( 1,{600},{000},{000}^{40000} \) (E) \( {4000}^{4,{000},{000}} \)
解答:(A)。
\( {40000} \times {40000}^{40000} = {40000}^{1} \times {40000}^{40000} = {40000}^{{40000} + 1} = {40000}^{40001} = \) \( {\left( {200}^{2}\right) }^{40001} = {\left( {200}\right) }^{80002} \)
例19. 下列哪一项与比值相同
\( \frac{{3}^{2017} \times {5}^{2019}}{{15}^{2018}}? \)
(A) \( 5/6 \) (B) \( 6/5 \) (C) \( 5/3 \) (D) \( 2/3 \) (E) \( 3/2 \)
解答:(C)。
\( \frac{{3}^{2017} \times {5}^{2019}}{{15}^{2018}} = \frac{{3}^{2017} \times {5}^{2019}}{{\left( 3 \times 5\right) }^{2018}} = \frac{{3}^{2017} \times {5}^{2019}}{{3}^{2018} \times {5}^{2018}} = \frac{5}{3}. \)
例20. 比值 \( \frac{{20}^{2019} - {20}^{2017}}{{20}^{20178} - {20}^{2016}} \) 与下列哪个
数相同？
(A)20(B)19(C)4(D)8(E)10
解答:(A)。
\[ \frac{{20}^{2019} - {20}^{2017}}{{20}^{2018} - {20}^{2016}} = \frac{{20}^{2017}\left( {{20}^{2} - 1}\right) }{{20}^{2016}\left( {{20}^{2} - 1}\right) } = {20}. \]
例21. 分数 \( \frac{{\left( {3}^{2018}\right) }^{2} - {\left( {3}^{2016}\right) }^{2}}{{\left( {3}^{2017}\right) }^{2} - {\left( {3}^{2015}\right) }^{2}} \) 化简后等于下列哪一项？
(A) 9 (B) \( 9/2 \) (C) \( 2/9 \) (D) 3 (E) 0
解答:(A)。
\[ \frac{{\left( {3}^{2018}\right) }^{2} - {\left( {3}^{2016}\right) }^{2}}{{\left( {3}^{2017}\right) }^{2} - {\left( {3}^{2015}\right) }^{2}} = \frac{\left( {{3}^{2018} - {3}^{2016}}\right) (\left( {{3}^{2018} + {3}^{2016}}\right) }{\left( {{3}^{2017} - {3}^{2015}}\right) (\left( {{3}^{2017} + {3}^{2015}}\right) } = \]
\[ \frac{{3}^{2015}\left( {{3}^{3} - 3}\right) \times {3}^{2015}\left( {{3}^{3} + 3}\right) }{{3}^{2015}\left( {{3}^{2} - 1}\right) \times {3}^{2015}\left( {{3}^{2} + 1}\right) } = \frac{\left( {{27} - 3}\right) \left( {{27} + 3}\right) }{\left( {9 - 1}\right) \left( {9 + 1}\right) } = \frac{{24} \times {30}}{8 \times {10}} = 3 \times 3 = 9 \]
例22.(2003 AMC 10 A)下列哪一项与
表达式 \( \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}} \) 相同？
(A) \( \sqrt{x} \) (B) \( \sqrt[3]{{x}^{2}} \) (C) \( \sqrt[3]{{x}^{27}} \) (D) \( \sqrt[3]{{x}^{54}} \) (E) \( \sqrt[8]{{x}^{80}} \)
解答:(A)。
方法一:
\[ \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}} = {\left( x{\left( x{\left( x \cdot {x}^{\frac{1}{2}}\right) }^{\frac{1}{3}}\right) }^{\frac{1}{3}}\right) }^{\frac{1}{3}} = {\left( x{\left( x{\left( {x}^{\frac{1}{2}}\right) }^{\frac{1}{3}}\right) }^{\frac{1}{3}}\right) }^{\frac{1}{3}} = {\left( x{\left( x{\left( {x}^{\frac{1}{2}}\right) }^{\frac{1}{3}}\right) }^{\frac{1}{3}}\right) }^{\frac{1}{3}} \]
\[ = {\left( x{\left( {x}^{\frac{3}{2}}\right) }^{\frac{1}{3}}\right) }^{\frac{1}{3}} = \left( {x{\left( {x}^{\frac{1}{2}}\right) }^{\frac{1}{3}} = \left( {x{\left( {x}^{\frac{3}{2}}\right) }^{\frac{1}{3}} = {\left( {x}^{\frac{3}{2}}\right) }^{\frac{1}{3}} = {x}^{\frac{1}{2}} = \sqrt{x}}\right. }\right. \]
方法二(我们的解法):
\[ \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}} = \sqrt[3]{\sqrt[3]{{x}^{3} \cdot x \cdot \sqrt[3]{x\sqrt{x}}}} = \sqrt[3]{\sqrt[3]{{x}^{4} \cdot \sqrt[3]{x\sqrt{x}}}} = \sqrt[3]{\sqrt[3]{\sqrt[3]{{x}^{12} \cdot x\sqrt{x}}}} = \]
\[ \sqrt[3]{\sqrt[3]{\sqrt[3]{{x}^{13}\sqrt{x}}}} = \sqrt[3]{\sqrt[3]{\sqrt[3]{\sqrt{{x}^{27}}}}} = {\left( {x}^{27}\right) }^{\frac{1}{2}\frac{1}{3}\frac{1}{3}\frac{1}{3}} = {\left( {x}^{27}\right) }^{\frac{1}{2} \cdot {27}} = {x}^{\frac{1}{2}} = \sqrt{x} \]
例23.(2008 AMC10 B)假设 \( x \) 为正实数。下列哪一项与 \( \sqrt[3]{x\sqrt{x}} \) 等价？
(A) \( {x}^{\frac{1}{6}} \) (B) \( {x}^{\frac{1}{4}} \) (C) \( {x}^{\frac{3}{8}} \) (D) \( {x}^{\frac{1}{2}} \) (E) \( x \)
解答:(D):
方法一(官方解答):
指数的性质表明
\[ \sqrt[3]{x\sqrt{x}} = {\left( x \cdot {x}^{\frac{1}{2}}\right) }^{\frac{1}{3}} = {\left( {x}^{\frac{2}{3}}\right) }^{\frac{1}{3}} = {x}^{\frac{1}{2}} \]
方法二(我们的解法):
\[ \sqrt[3]{x\sqrt{x}} = \sqrt[3]{\sqrt{{x}^{2} \cdot x}} = \sqrt[3]{\sqrt{{x}^{3}}} = \sqrt[{3 \times 2}]{{x}^{3}} = \sqrt{x} = {x}^{\frac{1}{2}} \]
例24. 假设 \( x < 0 \) 。下列哪一项与
\( \sqrt[4]{\frac{{x}^{3}}{1 - \frac{x - 1}{x}}}? \)
(A) \( - x \) (B) \( x \) (C) 1 (D \( \sqrt[4]{\frac{x}{2}} \) (E) \( x\sqrt{-1} \)
答案:(A)。
\[ \sqrt[4]{\frac{{x}^{3}}{1 - \frac{x - 1}{x}}} = \sqrt[4]{\frac{{x}^{3}}{\frac{x}{x} - \frac{x - 1}{x}}} = \sqrt[4]{\frac{{x}^{3}}{x - \left( {x - 1}\right) }} = \sqrt[4]{\frac{\frac{{x}^{3}}{1}}{\frac{1}{x}}} = \sqrt[4]{{x}^{4}} = - x. \]
题目
问题1. 若 \( m + \frac{1}{m} = 3 \) ，求 \( {m}^{4} + \frac{1}{{m}^{4}} \) 。
(A) 47 (B) 49 (C) 51 (D) 90 (E) 81
问题2: \( \operatorname{Find}{r}^{3} + \frac{1}{{r}^{3}} \) 当且仅当 \( r + \frac{1}{r} = \sqrt{2} \) 。
(A) 4 (B) \( \sqrt{2} \) (C) \( - \sqrt{2} \) (D) 2 (E) 1
问题3. 若 \( x = \frac{3 + \sqrt{13}}{2} \) ，求 \( {x}^{2} + \frac{1}{{x}^{2}} \) 的值。
(A) 13 (B) 14 (C) 10 (D) 12 (E) 11
问题4. 化简 \( \frac{\frac{{x}^{2}}{y} + \frac{{y}^{2}}{x}}{{y}^{2} - {xy} + {x}^{2}} \) 。
(A) \( {x}^{2} - {y}^{2} \) (B) \( \frac{xy}{x - y} \) (C) \( \frac{x + y}{xy} \) (D) \( x - y \) (E) 以上都不是
问题5. 下列哪一项与
\( \frac{3 - 6 + 9 - {12} + {15} - {18} + {21} - {24}}{2 - 4 + 6 - 8 + {10} - {12} + {14} - {16}}? \)
(A) -1 (B) \( - \frac{2}{3} \) (C) \( \frac{2}{3} \) (D) 1 (E) \( \frac{3}{2} \)
问题6. \( \frac{1 + 3 + 5 + 7}{2 + 4 + 6 + 8} - \frac{2 + 4 + 6 + 8}{1 + 3 + 5 + 7} \) 是多少？
(A) -1 (B) \( - \frac{5}{36} \) (C) \( - \frac{9}{20} \) (D) \( \frac{9}{20} \) (E) \( \frac{34}{3} \)
问题7. 下列哪一项等于 \( \frac{\frac{1}{4} - \frac{1}{5}}{\frac{1}{5} - \frac{1}{6}} \) ？
(A) \( \frac{1}{3} \) (B) \( \frac{1}{4} \) (C) \( \frac{2}{3} \) (D) \( \frac{3}{2} \) (E) \( \frac{4}{3} \)
问题8. 化简: \( \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}} \) 。
(A) \( \frac{5}{12} \) (B) \( \frac{3}{2} \) (C) \( \frac{5}{13} \) (D) \( \frac{5}{4} \) (E) 3
问题9. 下列哪一项等于该乘积
\( \frac{8}{4} \times \frac{12}{8} \times \frac{16}{12} \times \cdots \times \frac{{4n} + 4}{4n}\cdots \times \frac{2020}{2016}? \)
(A) 505 (B) 1346 (C) 2019 (D) 2016 (E) 4038
问题10. 对于非零数 \( a, b \) 和 \( c \) ，定义 \( \left( {a, b, c}\right) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \) 。
(6,36,27)等于多少？
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8
问题11. 对于正整数 \( a, b \) 和 \( c \) ，若 \( \left( {a, b, c}\right) = \frac{abc}{a + b + c} \) ，求(3,6,9)。
(A) 4 (B) 5 (C) 6 (D) 9 (E) 8
问题12. 求下列无穷级数前2006项的和: \( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots + \frac{1}{\left( {{2n} - 1}\right) \left( {{2n} + 1}\right) } + \cdots \) B. \( \frac{1}{16096143} \) C. \( \frac{2006}{16096143} \) D. \( \frac{2006}{4013} \) E. \( \frac{2006}{4011} \)
问题13. 计算 \( \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \cdots + \frac{1}{\left( {{3n} - 2}\right) \cdot \left( {{3n} + 1}\right) } \) 。
A. \( \frac{n}{\left( 3n + 1\right) } \) B. \( \frac{n}{\left( 3n - 1\right) } \) C. \( \frac{2n}{\left( 3n + 1\right) } \) D. \( \frac{n}{\left( n + 1\right) } \) E. \( \frac{n}{\left( n - 1\right) } \)
问题14. (2011 NC Algebra II) 设 \( f\left( x\right) = \frac{1}{x\left( {x + 1}\right) } \) 。求 \( f\left( 1\right) + f\left( 2\right) + \ldots \)
\( f\left( {2011}\right) \) .
(A) 1005/2011 (B) 2010/2011 (C) 2011/2012 (D) \( {4047}/{2012} \) (E) 6066/2011
问题15. 求乘积 \( \left( {1 - \frac{1}{2}}\right) \left( {1 - \frac{1}{3}}\right) \left( {1 - \frac{1}{4}}\right) \cdots \left( {1 - \frac{1}{2008}}\right) \) 。
(A) \( \frac{1}{{2008}!} \) (B) \( \frac{1}{2007} \) (C) \( \frac{{2007}!}{{2}^{2008}} \) (D) \( \frac{1}{2008} \) \( \epsilon \) 以上都不是
问题16. 求该连分数的值: \( 1 + \frac{2}{2 + \frac{2}{2 + \frac{2}{2 + \frac{2}{.}}}} \) 。
(A) \( \sqrt{5} \) (B) \( \frac{1 + \sqrt{5}}{2} \) (C) \( \sqrt{3} \) (D) \( \frac{1 + \sqrt{3}}{2} \) (E) \( \frac{1 - \sqrt{5}}{2} \) 。
问题17. 若 \( \frac{x}{y} = 3 \) ，求 \( \frac{4{x}^{3} + 3{x}^{2}y - 4{y}^{3}}{2{x}^{3} - {2x}{y}^{2} - 2{y}^{3}} \) 。
(A) \( \frac{131}{46} \) (B) \( \frac{108}{43} \) (C) \( \frac{27}{2} \) (D) \( \frac{4}{3} \) (E) \( \frac{3}{5} \)
问题18. \( {3000}\left( {3000}^{3000}\right) = \)
(A) \( {3000}^{3001} \) (B) \( {6000}^{3000} \) (C) \( {3000}^{6000} \) (D) \( 6,{000},{000}^{3000} \) (E) \( {3000}^{6,{000},{000}} \)
问题19. (2002 AMC 10 B) 下列哪一项与比值 \( \frac{{2}^{2001} \times {3}^{2003}}{{6}^{2002}}? \) 相同？
(A) \( 1/6 \) (B) \( 1/3 \) (C) \( 1/2 \) (D) \( 2/3 \) (E) \( 3/2 \)
问题20. 比值 \( \frac{{20}^{2019} - {20}^{2017}}{{20}^{2018} + {20}^{2017}} \) 最接近下列哪个
数？
(A) 20 (B) 19 (C) 4 (D) 8 (E) 10
问题21. (2013 AMC 10 A) \( \frac{{2}^{2014} + {2}^{2012}}{{2}^{2014} - {20}^{2012}} \) 的值是多少？
(A) -1 (B) 1 (C) \( \frac{5}{3} \) (D) 2012 (E) \( {2}^{4024} \)
问题22. 化简 \( \sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}} \cdot N \) ，其中 \( \sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}} \cdot N \) 为大于1的正整数。
(A) \( \sqrt{N} \) (B) \( {N}^{\frac{13}{54}} \) (C) \( {N}^{\frac{23}{54}} \) (D) \( {N}^{\frac{13}{27}} \) (E) \( {N}^{\frac{8}{9}} \)
问题23. 化简: \( \sqrt[x]{\sqrt{\sqrt[3]{25}}} \) ，其中 \( x \) 为大于1的自然数。
(A) \( {25}^{3x} \) (B) \( {25}^{\frac{1}{3x}} \) (C) \( {5}^{\frac{1}{3x}} \) (D \( {5}^{3x} \) (E) 以上都不是
问题24. 化简: \( \sqrt{\frac{{2}^{x + 4} - 2\left( {2}^{x + 1}\right) }{2\left( {2}^{x + 3}\right) }} \) 。
A. \( \frac{3}{8} \) B. \( \frac{\sqrt{3}}{4} \) C. \( {2}^{x} \) D. \( \frac{x\sqrt{3}}{4} \) E. \( \frac{\sqrt{3}}{2} \)
解答
问题1. 解答:(C)。
由于 \( {\left( x + y\right) }^{2} = {x}^{2} + {2xy} + {y}^{2} \) ，我们有:
\[ {x}^{2} + {y}^{2} = {\left( x + y\right) }^{2} - {2xy} \tag{1} \]
在(1)中令 \( x = m \) 和 \( y = \frac{1}{m} \) : \( {m}^{2} + \frac{1}{{m}^{2}} = {\left( m + \frac{1}{m}\right) }^{2} - {2m} \times \frac{1}{m} = {3}^{2} - 2 = 7 \) 。
\[ {m}^{4} + \frac{1}{{m}^{4}} = {\left( {m}^{2} + \frac{1}{{m}^{2}}\right) }^{2} + 2{m}^{2} \times \frac{1}{{m}^{2}} = {7}^{2} + 2 = {51}. \]
问题2:解答:(C)。
我们知道 \( {r}^{3} + \frac{1}{{r}^{3}} = \left( {r + \frac{1}{r}}\right) \left( {{r}^{2} - 1 + \frac{1}{{r}^{2}}}\right) \) ，且 \( {\left( r + \frac{1}{r}\right) }^{2} = {r}^{2} + 2 + \frac{1}{{r}^{2}} = 2 \) 。
第二个方程给出 \( {r}^{2} + \frac{1}{{r}^{2}} = 0 \Rightarrow {r}^{2} - 1 + \frac{1}{{r}^{2}} = - 1 \) 。
因此， \( {r}^{3} + \frac{1}{{r}^{3}} = \left( {r + \frac{1}{r}}\right) \left( {-1}\right) = - \sqrt{2} \) 。
问题3. 解答:(E)。
\[ {x}^{2} + \frac{1}{{x}^{2}} = {x}^{2} - {2x}\frac{1}{x} + \frac{1}{{x}^{2}} + 2 = {\left( x - \frac{1}{x}\right) }^{2} + 2 = {3}^{2} + 2 = {11} \]
问题4. 解答:(C)。
\[ \frac{\frac{{x}^{2}}{y} + \frac{{y}^{2}}{x}}{{y}^{2} - {xy} + {x}^{2}} = \frac{\frac{{x}^{3} + {y}^{3}}{xy}}{{y}^{2} - {xy} + {x}^{2}} = \frac{\left( {x + y}\right) \left( {{x}^{2} - {xy} + {y}^{2}}\right) }{{xy}\left( {{y}^{2} - {xy} + {x}^{2}}\right) } = \frac{x + y}{xy}. \]
问题5. 解答:(E)。
\[ \frac{3 - 6 + 9 - {12} + {15} - {18} + {21} - {24}}{2 - 4 + 6 - 8 + {10} - {12} + {14} - {16}} = \frac{3\left( {1 - 2 + 3 - 4 + 5 - 6 + 7 - 8}\right) }{2\left( {1 - 2 + 3 - 4 + 5 - 6 + 7 - 8}\right) } = \frac{3}{2}. \]
问题6. 解答:(D)。
\[ \frac{2 + 4 + 6 + 8}{1 + 3 + 5 + 7} - \frac{1 + 3 + 5 + 7}{2 + 4 + 6 + 8} = \frac{20}{16} - \frac{16}{20} = \frac{5}{4} - \frac{4}{5} = \frac{{25} - {16}}{20} = \frac{9}{20}. \]
问题7. 解答:(D)。
\[ \frac{\frac{1}{4} - \frac{1}{5}}{\frac{1}{5} - \frac{1}{6}} = \frac{\frac{5 - 4}{20}}{\frac{6 - 5}{30}} = \frac{\frac{1}{20}}{\frac{1}{30}} = \frac{30}{20} = \frac{3}{2}. \]
问题8. 答案:(A)。
\[ \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}} = \frac{1}{2 + \frac{1}{\frac{5}{2}}} = \frac{1}{2 + \frac{2}{5}} = \frac{1}{\frac{12}{5}} = \frac{5}{12}. \]
问题9. 答案:(A)。
\[ \frac{8}{4} \times \frac{12}{8} \times \frac{16}{12} \times \cdots \times \frac{{4n} + 4}{4n}\cdots \times \frac{2020}{2016} \]
\[ = \frac{8}{4} \times \frac{12}{8} \times \frac{16}{12} \times \cdots \times \frac{{4n} + 4}{4n}\cdots \times \frac{2020}{2016} = \frac{2020}{4} = {505} \]
问题10. 答案:(C)。
\[ \left( {a, b, c}\right) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \Rightarrow \]
\[ \left( {6,{36},{27}}\right) = \frac{6}{36} + \frac{36}{27} + \frac{27}{6} = \frac{1}{6} + \frac{27}{6} + \frac{4}{3} = \frac{14}{3} + \frac{4}{3} = \frac{18}{3} = 6 \]
问题11. 答案:(D)。
\[ \left( {a, b, c}\right) = \frac{abc}{a + b + c}\; \Rightarrow \;\left( {3,6,9}\right) = \frac{3 \times 6 \times 9}{3 + 6 + 9} = \frac{162}{18} = 9 \]
问题12. 答案:(D)。
\[ \text{By}\frac{1}{n\left( {n + k}\right) } = \frac{1}{k}\left( {\frac{1}{n} - \frac{1}{n + k}}\right) ,\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots + \frac{1}{\left( {{2n} - 1}\right) \left( {{2n} + 1}\right) } + \cdots \]
\[ = \frac{1}{2}\left( {\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5}\cdots + \frac{1}{4011} - \frac{1}{4013}}\right) = \frac{1}{2}\left( {\frac{1}{1} - \frac{1}{4013}}\right) = \frac{1}{2}\left( \frac{4012}{4013}\right) = \frac{2006}{4013}. \]
问题13. 答案:(A)。
\[ \text{By}\frac{1}{n\left( {n + k}\right) } = \frac{1}{k}\left( {\frac{1}{n} - \frac{1}{n + k}}\right) ,\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \cdots + \frac{1}{\left( {{3n} - 2}\right) \cdot \left( {{3n} + 1}\right) } \]
\[ = \frac{1}{3}\left( {\frac{1}{1} - \frac{1}{4} + \frac{1}{4} - \frac{1}{7}\cdots + \frac{1}{{3n} - 2} - \frac{1}{{3n} + 1}}\right) = \frac{1}{3}\left( {\frac{1}{1} - \frac{1}{{3n} + 1}}\right) = \frac{1}{3}\left( \frac{{3n} + 1 - 1}{{3n} + 1}\right) = \frac{n}{{3n} + 1} \]
问题14. 答案:(C)。
\[ \text{By}\frac{1}{n\left( {n + 1}\right) } = \frac{1}{n} - \frac{1}{n + 1}, f\left( 1\right) + f\left( 2\right) + \ldots f\left( {2011}\right) \]
\[ = \left. { = \frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5}\cdots + \frac{1}{2011} - \frac{1}{2012}}\right) = \frac{1}{1} - \frac{1}{2012}) = \frac{2011}{2012}\text{.} \]
问题15. 答案:(D)。
\[ \left( {1 - \frac{1}{2}}\right) \left( {1 - \frac{1}{3}}\right) \left( {1 - \frac{1}{4}}\right) \cdots \left( {1 - \frac{1}{2008}}\right) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\cdots \times \frac{2007}{2008} = \frac{1}{2008}. \]
问题16. 答案:(C)。
\[ \text{Let}\frac{2}{2 + \frac{2}{2 + \frac{2}{2 + \frac{2}{2}}}} = x\; \Rightarrow \;\frac{2}{2 + x} = x\; \Rightarrow \;{x}^{2} + {2x} - 2 = 0 \]
\[ \Rightarrow \;{x}^{2} + {2x} + 1 - 3 = 0\; \Rightarrow \;{\left( x + 1\right) }^{2} = 3\; \Rightarrow \;x = \sqrt{3} - 1 \]
\[ \text{ So }1 + \frac{2}{2 + \frac{2}{2 + \frac{2}{2 + \frac{2}{\ldots }}}} = 1 + \sqrt{3} - 1 = \sqrt{3}. \]
问题17. 答案:(A)。
由于已知 \( \frac{x}{y} = 3 \) ，因此我们知道 \( x = {3y} \) 。将其代入给定表达式，我们得到:
\[ \frac{4{x}^{3} + 3{x}^{2}y - 4{y}^{3}}{2{x}^{3} - {2x}{y}^{2} - 2{y}^{3}} = \frac{4 \times {\left( 3y\right) }^{3} + 3 \times {\left( 3y\right) }^{2}y - 4{y}^{3}}{2 \times {\left( 3y\right) }^{3} - 2 \times \left( {3y}\right) {y}^{2} - 2{y}^{3}} = \frac{{108}{y}^{3} + {27}{y}^{3} - 4{y}^{3}}{{54}{y}^{3} - 6{y}^{3} - 2{y}^{3}} = \frac{131}{46}. \]
问题18. 答案:(A)。
\[ {3000}\left( {30003000}\right) = \left( {3000}^{1}\right) \left( {3000}^{3000}\right) = {3000}^{1 + {3000}} = {3000}^{3001}. \]
所有其他选项均大于 \( {3000}^{3001} \) 。
问题19. 答案:(E)。
\[ \frac{{2}^{2001} \times {3}^{2003}}{{6}^{2002}} = \frac{{2}^{2001} \times {3}^{2003}}{{\left( 2 \times 3\right) }^{2002}} = \frac{{2}^{2001} \times {3}^{2003}}{{2}^{2002} \times {3}^{2002}} = \frac{3}{2}. \]
问题20. 答案:(B)。
\[ \frac{{20}^{2019} - {20}^{2017}}{{20}^{2018} + {20}^{2017}} = \frac{{20}^{2017}\left( {{20}^{2} - 1}\right) }{{20}^{2017}\left( {{20} + 1}\right) } = \frac{\left( {{20} + 1}\right) \left( {{20} - 1}\right) }{\left( {20} + 1\right) } = {19}. \]
问题21. 答案:(C)。
\[ \frac{{2}^{2014} + {2}^{2012}}{{2}^{2014} - {20}^{2012}} = \frac{{2}^{2012}\left( {{2}^{2} + 1}\right) }{{2}^{2012}\left( {{2}^{2} - 1}\right) } = \frac{5}{3}. \]
问题22。解答:(B)。
方法一:
\[ \sqrt{3\sqrt{N\sqrt[3]{N\sqrt[3]{N}}}} = \sqrt{3\sqrt{N\sqrt[3]{N \cdot {N}^{\frac{1}{3}}}}} = \sqrt{3\sqrt{N\sqrt[3]{{N}^{\frac{4}{3}}}}} = \sqrt{3\sqrt{N \cdot {N}^{\frac{4}{9}}}} = \sqrt{3\sqrt{\frac{13}{9}}} = {N}^{\frac{13}{54}}. \]
方法二:
\[ \sqrt{\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}}} = {\left( {\left( N{\left( N{\left( N\right) }^{\frac{1}{3}}\right) }^{\frac{1}{3}}\right) }^{\frac{1}{3}}\right) }^{\frac{1}{2}} = {\left( {\left( N\left( {N}^{\frac{1}{3}} \cdot {N}^{\frac{1}{9}}\right) \right) }^{\frac{1}{3}}\right) }^{\frac{1}{2}} = {\left( {N}^{\frac{1}{3}} \cdot {N}^{\frac{1}{9}} \cdot {N}^{\frac{1}{27}}\right) }^{\frac{1}{2}} = \]
\[ {N}^{\frac{13}{54}} \]
问题23。解答:(C)。
\[ \sqrt[x]{\sqrt{25}} = {25}^{\frac{1}{6x}} = {\left( {5}^{2}\right) }^{\frac{1}{6x}} = {5}^{\frac{1}{3x}} \]
问题24。解答:E。
\[ \sqrt{\frac{{2}^{x + 4} - 2\left( {2}^{x + 1}\right) }{2\left( {2}^{x + 3}\right) }} = \sqrt{\frac{{2}^{x + 4} - {2}^{x + 2}}{{2}^{x + 4}}} = \sqrt{\frac{{2}^{x + 4}\left( {1 - {2}^{-2}}\right) }{{2}^{x + 4}}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]