Lũy thừa của một số hữu tỉ

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

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

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

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

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

$\begin{array}{l}{2^n} + {2^{n + 3}} = 72\\{2^n} + {2^n}{.2^3} = 72\\{2^n}.\left( {1 + {2^3}} \right) = 72\\{2^n}.9 = 72\\{2^n} = 72:9\\{2^n} = 8\\{2^n} = {2^3}\\n = 3.\end{array}$

Vậy $n=3$.

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

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

Vậy $S{\rm{ }} = {\rm{ }}1155$.

Ta thấy $1155$ chia hết cho cả $3$ và $5$.

$\begin{array}{l}{3^n} + {3^{n + 2}} = 90\\{3^n} + {3^n}{.3^2} = 90\\{3^n}.\left( {1 + {3^2}} \right) = 90\\{3^n}.10 = 90\\{3^n} = 90:10\\{3^n} = 9\\{3^n} = {3^2}\\n = 2.\end{array}$

Vậy $n=2$.

$\begin{array}{l}{7^{2n}} + {7^{2n + 2}} = 2450\\{7^{2n}} + {7^{2n}}{.7^2} = 2450\\{7^{2n}}.\left( {1 + {7^2}} \right) = 2450\\{7^{2n}}.50 = 2450\\{7^{2n}} = 2450:50\\{7^{2n}} = 49\\{7^{2n}} = {7^2}\\2n = 2\\n = 1.\end{array}$.

Vậy $n=1$

Ta có: ${\left( {x + \dfrac{1}{3}} \right)^2} \ge 0$ với mọi $x$

$\Rightarrow {\left( {x + \dfrac{1}{3}} \right)^2} + \dfrac{1}{{100}} \ge 0+ \dfrac{1}{{100}}$

$\Rightarrow {\left( {x + \dfrac{1}{3}} \right)^2} + \dfrac{1}{{100}} \ge \dfrac{1}{{100}}$

Do đó GTNN biểu thức đạt được là $\dfrac{1}{{100}}$ khi và chỉ khi

$(x + \dfrac{1}{3})^2 = 0$ $\Rightarrow x + \dfrac{1}{3} = 0$ hay $x = - \dfrac{1}{3}$.

Vậy giá trị nhỏ nhất là $\dfrac{1}{100}$ tại $x = \dfrac{{ - 1}}{3}$.

Ta có: ${\left( {2x - 0,5} \right)^4} \ge 0$$\Rightarrow {\left( {2x - 0,5} \right)^4} + \dfrac{{10}}{{11}} \ge \dfrac{{10}}{{11}}$ với mọi $x$.

Dấu “=” xảy ra khi $2x - 0,5 = 0$ $\Rightarrow 2x - \dfrac{1}{2} = 0 \Rightarrow 2x = \dfrac{1}{2} \Rightarrow x = \dfrac{1}{2}:2 \Rightarrow x = \dfrac{1}{4}$.

Do đó giá trị nhỏ nhất của biểu thức là: $\dfrac{{10}}{{11}}$ khi $x = \dfrac{1}{4}$.

$\begin{array}{l}x:{\left( {\dfrac{2}{3}} \right)^2} = {\left( {\dfrac{2}{3}} \right)^4}\\x = {\left( {\dfrac{2}{3}} \right)^4}.{\left( {\dfrac{2}{3}} \right)^2}\\x = {\left( {\dfrac{2}{3}} \right)^6}\end{array}$

Vậy $x = {\left( {\dfrac{2}{3}} \right)^6}$.

$\begin{array}{l}x:{\left( {\dfrac{3}{5}} \right)^3} = {\left( {\dfrac{3}{5}} \right)^5}\\x = {\left( {\dfrac{3}{5}} \right)^5}.{\left( {\dfrac{3}{5}} \right)^3}\\x = {\left( {\dfrac{3}{5}} \right)^8}\end{array}$

Vậy $x = {\left( {\dfrac{3}{5}} \right)^8}$.

${\left( {\dfrac{{ - 1}}{9}} \right)^3}.{\left( { - 9} \right)^3} = \dfrac{{{{( - 1)}^3}}}{{{9^3}}}.{\left( { - 9} \right)^3} = \dfrac{{{{\left( { - 1} \right)}^3}.{{\left( { - 9} \right)}^3}}}{{{9^3}}} = \dfrac{{\left( { - 1} \right).( - 1){{.9}^3}}}{{{9^3}}} = ( - 1).( - 1) = 1$

Ta có $A=\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}$

Ta thấy $A = \dfrac{4}{5} < \dfrac{5}{5} = 1$

Vậy $A < 1$.

${\left( { - 2} \right)^4}.{\left( {\dfrac{1}{2}} \right)^4} = {\left( { - 2} \right)^4}.\dfrac{{{{\left( 1 \right)}^4}}}{{{{\left( 2 \right)}^4}}} = \dfrac{{{{\left( { - 2} \right)}^4}.1}}{{{2^4}}} = \dfrac{{{2^4}}}{{{2^4}}} = 1$.

Ta có: $\dfrac{{{4^3}{{.2}^5} + {8^2}}}{{{8^3}.3 + {{16.3}^2}}} = \dfrac{{{{({2^2})}^3}{{.2}^5} + {{({2^3})}^2}}}{{{{({2^3})}^3}.3 + {2^4}{{.3}^2}}} = \dfrac{{{2^6}{{.2}^5} + {2^6}}}{{{2^9}.3 + {2^4}{{.3}^2}}}$ $= \dfrac{{{2^{11}} + {2^6}}}{{{2^9}.3 + {2^4}{{.3}^2}}}$

$= \dfrac{{{2^4}{{.2}^7} + {2^4}{{.2}^2}}}{{{2^4}{{.2}^5}.3 + {2^4}{{.3}^2}}}$ $= \dfrac{{{2^4}({2^7} + {2^2})}}{{{2^4}({2^5}.3 + {3^2})}}$ $= \dfrac{{{2^7} + {2^2}}}{{{2^5}.3 + {3^2}}} = \dfrac{{132}}{{105}} = \dfrac{{44}}{{35}}$.

Ta thấy $A = \dfrac{{44}}{{35}} > \dfrac{{35}}{{35}} = 1$

Vậy $A > 1$.

$\begin{array}{l}{3^x} = {\left( {{3^2}} \right)^4}\\{3^x} = {3^8}\\x = 8\end{array}$

Vậy $x = 8$.

$\begin{array}{l}{9^x} = {\left( {{9^3}} \right)^2}\\{9^x} = {9^{3.2}}\\{9^x} = {9^6}\\x = 6\end{array}$

Vậy $x = 6$.

$\begin{array}{l}{4^x} = {\left( {{4^3}} \right)^7}\\{4^x} = {4^{21}}\\x = 21\end{array}$

Vậy $x = 21$.

${( - 3)^5}.{\left( {\dfrac{1}{3}} \right)^5} = {( - 3)^5}.\dfrac{{{{(1)}^5}}}{{{3^5}}} = \dfrac{{{{\left( { - 3} \right)}^5}.1}}{{{3^5}}} = \dfrac{{ - {3^5}.1}}{{{3^5}}} = - 1$

Ta có: ${4^{24}} = {4^{2.12}} = {\left( {{4^2}} \right)^{12}} = {16^{12}}$

Ta có: ${2^{18}} = {\left( {{2^3}} \right)^6} = {8^6}$

Ta có: ${5^{16}} = {5^{2.8}} = {\left( {{5^2}} \right)^8} = {25^8}$