Lũy thừa của một số hữu tỉ

Ta có \({\left( {\dfrac{2}{3}} \right)^3}\)\( = \dfrac{{{2^3}}}{{{3^3}}} = \dfrac{8}{{27}}\)

Ta có \({x^1} = x;\)\({x^0} = 1\)\(\left( {x \ne 0} \right)\) nên A, B, C sai

${\left( {\dfrac{x}{y}} \right)^n} = \dfrac{{{x^n}}}{y^n}\left( {y \ne 0;\,n \in \mathbb{N}} \right)$ nên D đúng.

Ta có \({\left( {\dfrac{1}{7}} \right)^2}{.7^2} = \dfrac{1}{{{7^2}}}{.7^2} = \dfrac{{{7^2}}}{{{7^2}}} = 1\)

Ta có ${\left( {-2019} \right)^0} = 1$ nên A đúng.

+)  ${4^6}:{\rm{ }}{4^4} = {4^2} = 16$ nên C đúng

+)  ${\left( {-3} \right)^3}.{\left( {-{\rm{ }}3} \right)^{{\rm{ }}2}} = {\left( { - 3} \right)^{3 + 2}} = {\left( { - 3} \right)^5}$nên D đúng

+) $\left( {0,5} \right).{\left( {0,5} \right)^2} = {\left( {0,5} \right)^3} = {\left( {\dfrac{1}{2}} \right)^3} = \dfrac{1}{8}$ nên B sai.

Ta có

+) ${x^{18}}:{x^6} = {x^{12 - 6}} = {x^{12}}(x\; \ne 0)$ nên A đúng.

+) ${x^4}.{\rm{ }}{x^8} = {x^{4 + 8}} = {x^{12}}$ nên B đúng.

+ ${\left( {{x^3}} \right)^4} = {x^{3.4}} = {x^{12}}$ nên D đúng.

Ta thấy ở đáp án C: \({x^2}.{x^6} = {x^{2 + 6}} = {x^8} \ne {x^{12}}\)

nên C sai.

Ta có: \({2^{24}} = {2^{3.8}} = {\left( {{2^3}} \right)^8} = {8^8}\)

${2^x}\; = {\left( {{2^2}} \right)^5} \Leftrightarrow {2^x} = {2^{2.5}} \Leftrightarrow {2^x} = {2^{10}} \Leftrightarrow x = 10$

$a{\rm{ }}:{\left( {\dfrac{1}{3}} \right)^4} = {\left( {\dfrac{1}{3}} \right)^3}$

$a = {\left( {\dfrac{1}{3}} \right)^3}.{\left( {\dfrac{1}{3}} \right)^4}$

$a = {\left( {\dfrac{1}{3}} \right)^{3 + 4}}$

$a = {\left( {\dfrac{1}{3}} \right)^7}$

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 cần tìm là $\dfrac{1}{100}.$

\({20^n}\;:\;{5^n} = 4\)  

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

\({4^n} = 4\)

\(n = 1\)

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

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

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

Vậy $S{\rm{ }} = {\rm{ }}1155$.

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

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

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

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

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

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

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.