Lũy thừa với số mũ tự nhiên của một số hữu tỉ

\({20^n}\;:\;{5^n} = 4\)  

\({(20:5)^n} = 4\)

\({4^n} = 4\)

\(n = 1\)

$A = \dfrac{{{2^7}{{.9}^3}}}{{{6^5}{{.8}^2}}} = \dfrac{{{2^7}.{{\left( {{3^2}} \right)}^3}}}{{{2^5}{{.3}^5}.{{\left( {{2^3}} \right)}^2}}} = \dfrac{{{2^7}{{.3}^6}}}{{{2^5}{{.2}^6}{{.3}^5}}}$$ = \dfrac{{{2^7}{{.3}^6}}}{{{2^{11}}{{.3}^5}}} = \dfrac{{1.3}}{{{2^4}.1}} = \dfrac{3}{{16}}$

Ta có \(\dfrac{{{4^6}{{.9}^5} + {6^9}.120}}{{{8^4}{{.3}^{12}} - {6^{11}}}} = \dfrac{{{{\left( {{2^2}} \right)}^6}.{{\left( {{3^2}} \right)}^5} + {6^9}.120}}{{{{\left( {{2^3}} \right)}^4}{{.3}^{12}} - {6^{11}}}}\)\( = \dfrac{{{2^{12}}{{.3}^{10}} + {6^9}.6.20}}{{{2^{12}}{{.3}^{12}} - {6^{11}}}} = \dfrac{{{2^2}{{.2}^{10}}{{.3}^{10}} + {6^{10}}.20}}{{{{\left( {2.3} \right)}^{12}} - {6^{11}}}}\)\( = \dfrac{{{2^2}{{.6}^{10}} + {6^{10}}.20}}{{{6^{12}} - {6^{11}}}}\)\( = \dfrac{{{6^{10}}\left( {{2^2} + 20} \right)}}{{{6^{10}}\left( {{6^2} - 6} \right)}} = \dfrac{{24}}{{30}} = \dfrac{4}{5}\)

\({\left( {5x - 1} \right)^6} = 729\)

\({\left( {5x - 1} \right)^6} = {(3)^6}\)

Trường hợp 1:

$\begin{array}{l}5x-1 = 3\\5x = 4\\x = \dfrac{4}{5}\end{array}$

Trường hợp 2:

$\begin{array}{l}5x-1 =  - 3\\5x =  - 2\\x =  - \dfrac{2}{5}\end{array}$

Vậy \(x = \dfrac{4}{5}\) hoặc \(x =  - \dfrac{2}{5}\)

\({\left( {2x + 1} \right)^3} =  - {0,1^3} = {\left( { - 0,1} \right)^3}\)

\(2x + 1 =  - 0,1\)

\(2x =  - 0,1 - 1\)

\(2x =  - 1,1\)

\(x =  - 1,1:2\)

\(x =  - 0,55\)  

Vậy $x =  - 0,55$.

\({5^n} + {5^{n + 2}} = 650\)

\({5^n} + {5^n}{.5^2} = 650\)

\({5^n}\left( {1 + {5^2}} \right) = 650\)

\({5^n}\left( {1 + 25} \right) = 650\)

\({5^n}.26 = 650\)

\({5^n} = 650:26\)

\({5^n} = 25\)

\({5^n} = {5^2}\)

\(n = 2\)

Vậy $n = 2$

Ta có: \({1^2} + {2^2} + {3^2} + ... + {10^2} = 385\)

Suy ra \({1^2} + {3^2} + {5^2} + {7^2} + {9^2}\)\( = 385 - \left( {{2^2} + {4^2} + {6^2} + {8^2} + {{10}^2}} \right)\)\( = 385 - {2^2}\left( {{1^2} + {2^2} + {3^2} + {4^2} + {5^2}} \right)\)

Và \({12^2} + {14^2} + {16^2} + {18^2} + {20^2} \)\(= {2^2}.\left( {{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}} \right)\)

Suy ra \(S = {2^2}.({6^2} + {7^2} + {8^2} + {9^2} + {{10}^2})\)\( - 385 + {2^2}({1^2} + {2^2} + {3^2} + {4^2} + {5^2})\)

 \(S = 2^2.( {1^2} + {2^2} + {3^2} + {4^2} + {5^2} + {6^2} \)\(+ {7^2} + {8^2} + {9^2} + {10}^2 ) - 385 \)\(= 4.385 - 385 = 1155\)

Vậy $S = 1155$.

\(A = 1 - \dfrac{3}{4} + {\left( {\dfrac{3}{4}} \right)^2} - {\left( {\dfrac{3}{4}} \right)^3} + {\left( {\dfrac{3}{4}} \right)^4} - ... - {\left( {\dfrac{3}{4}} \right)^{2017}} + {\left( {\dfrac{3}{4}} \right)^{2018}}\)

\( \Rightarrow \dfrac{3}{4}A = \dfrac{3}{4} - {\left( {\dfrac{3}{4}} \right)^2} + {\left( {\dfrac{3}{4}} \right)^3} - {\left( {\dfrac{3}{4}} \right)^4} + ...\) \( + {\left( {\dfrac{3}{4}} \right)^{2017}} - {\left( {\dfrac{3}{4}} \right)^{2018}} + {\left( {\dfrac{3}{4}} \right)^{2019}}\)  

\( \Rightarrow A + \dfrac{3}{4}A = 1 + {\left( {\dfrac{3}{4}} \right)^{2019}}\)

\( \Rightarrow \left( {1 + \dfrac{3}{4}} \right)A = 1 + {\left( {\dfrac{3}{4}} \right)^{2019}}\)

\( \Rightarrow \dfrac{7}{4}.A = 1 + {\left( {\dfrac{3}{4}} \right)^{2019}}\)

\( \Rightarrow A = \left[ {1 + {{\left( {\dfrac{3}{4}} \right)}^{2019}}} \right]:\dfrac{7}{4} = \left[ {1 + {{\left( {\dfrac{3}{4}} \right)}^{2019}}} \right].\dfrac{4}{7}\)

Suy ra \(A > 0\,\,\,\,\,\,\,\left( 1 \right)\)

Vì \({\left( {\dfrac{3}{4}} \right)^{2019}} < \dfrac{3}{4} \Rightarrow A < \left( {1 + \dfrac{3}{4}} \right).\dfrac{4}{7} = 1\,\,\,\,\,\,\,\left( 2 \right)\)

Từ \(\left( 1 \right)\) và \(\left( 2 \right)\) suy ra \(0 < A < 1\).

Vậy \(A\) không phải là số nguyên.