Các bài toán liên quan đến lũy thừa với số mũ hữu tỉ.

Ta có: ${x^m}.{y^n} \ne {\left( {xy} \right)^{m + n}} \Rightarrow$ công thức sai.

Ta có $K = {\left( {{x^{\frac{1}{2}}} - {y^{\frac{1}{2}}}} \right)^2}{\left( {1 - 2\sqrt {\dfrac{y}{x}}  + \dfrac{y}{x}} \right)^{ - 1}}$

$\Leftrightarrow K = \dfrac{{{{\left( {\sqrt x  - \sqrt y } \right)}^2}}}{{{{\left( {\sqrt {\dfrac{y}{x}}  - 1} \right)}^2}}} = \dfrac{{x{{\left( {\sqrt x  - \sqrt y } \right)}^2}}}{{{{\left( {\sqrt y  - \sqrt x } \right)}^2}}} = x.$

$\begin{array}{l}f\left( a \right) = \dfrac{{{a^{\frac{2}{3}}}\left( {\sqrt[3]{{{a^{ - 2}}}} - \sqrt[3]{a}} \right)}}{{{a^{\frac{1}{8}}}\left( {\sqrt[8]{{{a^3}}} - \sqrt[8]{{{a^{ - 1}}}}} \right)}} = \dfrac{{{a^{\frac{2}{3}}}\left( {{a^{\frac{{ - 2}}{3}}} - {a^{\frac{1}{3}}}} \right)}}{{{a^{\frac{1}{8}}}\left( {{a^{\frac{3}{8}}} - {a^{\frac{{ - 1}}{8}}}} \right)}} = \dfrac{{{a^0} - {a^1}}}{{{a^{\frac{1}{2}}} - {a^0}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{1 - a}}{{\sqrt a  - 1}} = \dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}{{\sqrt a  - 1}} =  - \left( {\sqrt a  + 1} \right).\end{array}$

Thay $a = {2019^{2018}}$ vào ta được $M = f\left( {{{2019}^{2018}}} \right) =  - \left( {\sqrt {{{2019}^{2018}}}  + 1} \right) =  - {2019^{1009}} - 1.$

Ta có: $A = \frac{{\sqrt[3]{{{a^7}}}.{a^{\frac{{11}}{3}}}}}{{{a^4}.\sqrt[7]{{{a^{ - 5}}}}}} = \frac{{{a^{\frac{7}{3}}}.{a^{\frac{{11}}{3}}}}}{{{a^4}.{a^{\frac{{ - 5}}{7}}}}} = {a^{\frac{7}{3} + \frac{{11}}{3} - 4 + \frac{5}{7}}} = {a^{\frac{{19}}{7}}}.$

$\Rightarrow \left\{ \begin{array}{l}m = 19\\n = 7\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{m^2} + {n^2} = 410\\{m^2} - {n^2} = 312\end{array} \right..$

$\begin{array}{l}\dfrac{{\sqrt[3]{{{a^2}\sqrt a }}}}{{{a^3}}} = \dfrac{{\sqrt[3]{{{a^2}.{a^{\frac{1}{2}}}}}}}{{{a^3}}} = \dfrac{{\sqrt[3]{{{a^{\frac{5}{2}}}}}}}{{{a^3}}} = \dfrac{{{a^{\frac{5}{6}}}}}{{{a^3}}} = {a^{\frac{5}{6} - 3}} = {a^{ - \frac{{13}}{6}}}\\ \Rightarrow \alpha  =  - \dfrac{{13}}{6} \Leftrightarrow \alpha  \in \left( { - 3; - 2} \right)\end{array}$

Ta có: $P = {a^{\frac{3}{2}}}.\sqrt[3]{a} = {a^{\frac{3}{2}}}.{a^{\frac{1}{3}}} = {a^{\frac{3}{2} + \frac{1}{3}}} = {a^{\frac{{11}}{6}}}$

$P = \dfrac{{\sqrt[5]{4}.\sqrt[4]{{64}}.{{(\sqrt[3]{{\sqrt 2 }})}^4}}}{{\sqrt[3]{{\sqrt[3]{{32}}}}}} = \dfrac{{{2^{\frac{2}{5}}}{{.2}^{\frac{6}{4}}}{{.2}^{\frac{4}{6}}}}}{{{2^{\frac{5}{9}}}}} = {2^{\frac{2}{5} + \frac{6}{4} + \frac{4}{6} - \frac{5}{9}}} = {2^{\frac{{181}}{{90}}}}$

Vậy $P = {2^{\frac{{181}}{{90}}}}.$

Ta có : $P = \dfrac{{{{125}^6}.{{\left( { - 16} \right)}^3}2.\left( { - {2^3}} \right)}}{{{{25}^3}.{{\left( { - {5^2}} \right)}^4}}} = \dfrac{{{5^{18}}{2^{12}}{{.2.2}^3}}}{{{5^6}{{.5}^{2.4}}}} = {5^4}{.2^{16}}$

Vậy $P = {5^4}{.2^{16}}$.

$P = \sqrt[5]{{{x^2}\sqrt[3]{x}}} = \sqrt[5]{{{x^2}.{x^{\frac{1}{3}}}}} = {\left( {{x^{2 + \frac{1}{3}}}} \right)^{\frac{1}{5}}}$$= {x^{\frac{7}{{15}}}}$

Vậy $P = {x^{\frac{7}{{15}}}}.$

$P = \dfrac{{\sqrt[5]{{{b^2}\sqrt b }}}}{{\sqrt[3]{{b\sqrt b }}}} = \dfrac{{\sqrt[5]{{{b^2}.{b^{\frac{1}{2}}}}}}}{{\sqrt[3]{{b.{b^{\frac{1}{2}}}}}}}$= $\dfrac{{\sqrt[5]{{{b^{\frac{5}{2}}}}}}}{{\sqrt[3]{{{b^{\frac{3}{2}}}}}}} = \dfrac{{{b^{\frac{5}{{2.5}}}}}}{{{b^{\frac{3}{{2.3}}}}}} =$ $\dfrac{{{b^{\frac{1}{2}}}}}{{{b^{\frac{1}{2}}}}} = 1$

Vậy $P = 1.$

Ta có:

$P = \left( {{a^{\dfrac{1}{4}}} - {b^{\dfrac{1}{4}}}} \right)\left( {{a^{\dfrac{1}{4}}} + {b^{\dfrac{1}{4}}}} \right)\left( {{a^{\dfrac{1}{2}}} + {b^{\dfrac{1}{2}}}} \right) = \left( {{a^{\dfrac{1}{2}}} - {b^{\dfrac{1}{2}}}} \right)\left( {{a^{\dfrac{1}{2}}} + {b^{\dfrac{1}{2}}}} \right) = a - b$

Vậy  $P = a - b$.

$\begin{array}{l}P = {\left( {2\sqrt 6  - 5} \right)^{2020}}{\left( {2\sqrt 6  + 5} \right)^{2021}}\\\,\,\,\,\, = {\left[ {\left( {2\sqrt 6  - 5} \right)\left( {2\sqrt 6  + 5} \right)} \right]^{2020}}.\left( {2\sqrt 6  + 5} \right)\\\,\,\,\, = {\left( {24 - 25} \right)^{2020}}.\left( {2\sqrt 6  + 5} \right) = 2\sqrt 6  + 5\end{array}$

$P = \left( {\sqrt {ab}  - \dfrac{{ab}}{{a + \sqrt {ab} }}} \right):\dfrac{{\sqrt[4]{{ab}} - \sqrt b }}{{a - b}}$ $= \left( {\dfrac{{\sqrt {ab} \left( {a + \sqrt {ab} } \right) - ab}}{{a + \sqrt {ab} }}} \right).\dfrac{{a - b}}{{\sqrt[4]{{ab}} - {{\left( {\sqrt[4]{b}} \right)}^2}}}$$= \dfrac{{a.\sqrt {ab}  + ab - ab}}{{{{\left( {\sqrt a } \right)}^2} + \sqrt a .\sqrt b }}.\dfrac{{\left( {\sqrt a  - \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)}}{{\sqrt[4]{b}\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)}}$

$= \dfrac{{a\sqrt {ab} }}{{\sqrt a \left( {\sqrt a  + \sqrt b } \right)}}.\dfrac{{\left( {\sqrt a  - \sqrt b } \right)\left( {\sqrt a  + \sqrt b } \right)}}{{\sqrt[4]{b}\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)}}$  $= \dfrac{{a\sqrt a .\sqrt b }}{{\sqrt a }}.\dfrac{{{{\left( {\sqrt[4]{a}} \right)}^2} - {{\left( {\sqrt[4]{b}} \right)}^2}}}{{\sqrt[4]{b}\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)}}$ $= \dfrac{{a\sqrt b .\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)\left( {\sqrt[4]{a} + \sqrt[4]{b}} \right)}}{{\sqrt[4]{b}\left( {\sqrt[4]{a} - \sqrt[4]{b}} \right)}}$ $= \dfrac{{a{{\left( {\sqrt[4]{b}} \right)}^2}.\left( {\sqrt[4]{a} + \sqrt[4]{b}} \right)}}{{\sqrt[4]{b}}} = a\sqrt[4]{b}\left( {\sqrt[4]{a} + \sqrt[4]{b}} \right)$

Vậy $P = a\sqrt[4]{b}(\sqrt[4]{a} + \sqrt[4]{b}).$

Ta có:

$\begin{array}{l}C = \dfrac{{{{\left( {{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right)}^2}}}{{\sqrt[3]{{ab}}}}:\left( {2 + \sqrt[3]{{\dfrac{a}{b}}} + \sqrt[3]{{\dfrac{b}{a}}}} \right) = \dfrac{{{{\left( {\sqrt[3]{a} + \sqrt[3]{b}} \right)}^2}}}{{\sqrt[3]{{ab}}}}:\left( {\dfrac{{2\sqrt[3]{{ab}} + \sqrt[3]{{{a^2}}} + \sqrt[3]{{{b^2}}}}}{{\sqrt[3]{{ab}}}}} \right)\\\,\,\,\,\, = \dfrac{{\sqrt[3]{{{a^2}}} + 2\sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}}}{{\sqrt[3]{{ab}}}}.\dfrac{{\sqrt[3]{{ab}}}}{{\sqrt[3]{{{a^2}}} + 2\sqrt[3]{{ab}} + \sqrt[3]{{{b^2}}}}} = 1.\end{array}$

Ta có: $a = {1^{3,8}} = 1$; $b = {2^{ - 1}} = \dfrac{1}{2} = 0,5$ và $c = {\left( {\dfrac{1}{2}} \right)^{ - 3}} = {2^3} = 8.$

Mà $0,5 < 1 < 8 \Rightarrow b < a < c$

Vì $0 < \sqrt 2  - 1 < 1$ nên ${\left( {\sqrt 2  - 1} \right)^m} < {\left( {\sqrt 2  - 1} \right)^n} \Leftrightarrow m > n$.

Đáp án A: Vì $a > b > 0$ và $2 > 0$ nên ${a^2} > {b^2}$ (A sai).

Đáp án B: Vì $a > 1$ và $- 2 >  - 3$ nên ${a^{ - 2}} > {a^{ - 3}}$ (B sai).

Đáp án C: Vì $a > b > 0$ và $- \dfrac{3}{2} < 0$ nên ${a^{ - \dfrac{3}{2}}} < {b^{ - \dfrac{3}{2}}}$ (C đúng).

Đáp án D: Vì $0 < b < 1$ và $- 2 >  - \dfrac{5}{2}$ nên ${b^{ - 2}} < {b^{ - \dfrac{5}{2}}}$  (D sai).

Vì $- \dfrac{1}{4} >  - \dfrac{1}{3}$ nên ${\left( {a - 2} \right)^{ - \dfrac{1}{4}}} \le {\left( {a - 2} \right)^{ - \dfrac{1}{3}}} \Leftrightarrow 0 < a - 2 \le 1 \Leftrightarrow 2 < a \le 3$.

Vì $\dfrac{3}{4} < 2$ nên ${\left( {2 - a} \right)^{\dfrac{3}{4}}} > {\left( {2 - a} \right)^2} \Leftrightarrow 0 < 2 - a < 1 \Leftrightarrow 1 < a < 2$.

$\begin{array}{l}\,\,\,\,\,\sqrt {a.\sqrt[3]{{a.\sqrt[4]{a}}}}  = \sqrt[{24}]{{{2^5}}}.\dfrac{1}{{\sqrt {{2^{ - 1}}} }}\\ \Leftrightarrow \sqrt {a.\sqrt[3]{{a.{a^{\frac{1}{4}}}}}}  = {2^{\frac{5}{{24}}}}{.2^{\frac{1}{2}}}\\ \Leftrightarrow \sqrt {a.\sqrt[3]{{{a^{\frac{5}{4}}}}}}  = {2^{\frac{{17}}{{24}}}}\\ \Leftrightarrow \sqrt {a.{a^{\frac{5}{{12}}}}}  = {2^{\frac{{17}}{{24}}}}\\ \Leftrightarrow \sqrt {{a^{\frac{{17}}{{12}}}}}  = {2^{\frac{{17}}{{24}}}}\\ \Leftrightarrow {a^{\frac{{17}}{{24}}}} = {2^{\frac{{17}}{{24}}}}\\ \Leftrightarrow a = 2\end{array}$