Bài 103868

Question

Viết dưới dạng lũy thừa với số mũ hữu tỉ các biểu thức sau :
$a)$ $\sqrt[5]{{2\sqrt[3]{{2\sqrt 2 }}}}$                    $b)$ $\sqrt {a\sqrt {a\sqrt {a\sqrt a } } } :{a^{\frac{{11}}{6}}}$ ($a > 0)$
$c)$ $\sqrt[4]{{{x^2}\sqrt[3]{x}}}$ $(x > 0)$         $d)$ $\sqrt[5]{{\frac{b}{a}\sqrt[a]{{\frac{a}{b}}}}}$    $(ab > 0)$

score 0 · Answer 1

$a)$ Ta có : $\sqrt[5]{{2\sqrt[3]{{2\sqrt 2 }}}}$    $ = {\left( {2\sqrt[2]{{2\sqrt 2 }}}
\right)^{\frac{1}{5}}} = {2^{\frac{1}{5}}}{\left( {\sqrt[3]{{2\sqrt 2 }}} \right)^{\frac{1}{5}}} =
{2^{\frac{1}{5}}}{\left[ {{{\left( {2\sqrt 2 } \right)}^{\frac{1}{3}}}} \right]^{\frac{1}{5}}}$
                $ = {2^{\frac{1}{2}}}{\left( {2\sqrt 2 }
\right)^{\frac{1}{{15}}}} = {2^{\frac{1}{2}}}{2^{\frac{3}{2}.\frac{1}{{15}}}} =
{2^{\frac{1}{2}}}{.2^{\frac{1}{{10}}}} = {2^{\frac{1}{2} + \frac{1}{{10}}}} =
{2^{\frac{3}{5}}}$
$b)$ $\sqrt {a\sqrt {a\sqrt {a\sqrt a } } } = {a^{\frac{1}{2}}}\sqrt[4]{{a\sqrt {a\sqrt a } }} =
{a^{\frac{1}{2}}}.{a^{\frac{1}{4}}}.\sqrt[8]{{a\sqrt a }} =
{a^{\frac{1}{2}}}{a^{\frac{1}{4}}}{a^{\frac{1}{8}}}{a^{\frac{1}{{16}}}}$
            $ = {a^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}}}}
= {a^{\frac{{15}}{{16}}}}$
$ \Rightarrow $ $\sqrt {a\sqrt {a\sqrt {a\sqrt a } } } :{a^{\frac{{11}}{6}}} =
{a^{\frac{{15}}{{16}}}}:{a^{\frac{{11}}{6}}} = {a^{\frac{{15}}{{16}} - \frac{{11}}{6}}} =
{a^{ - \frac{{43}}{{16}}}} = \frac{1}{{{a^{\frac{{43}}{{16}}}}}}$
$c)$ $\sqrt[4]{{{x^2}\sqrt[3]{x}}} = \sqrt[4]{{{x^2}}}.\sqrt[{12}]{{{x^2}}} =
{x^{\frac{1}{2}}}.{x^{\frac{1}{{12}}}} = {x^{\frac{1}{2} + \frac{1}{{12}}}} =
{x^{\frac{7}{{12}}}}$
$d)$ $\sqrt[5]{{\frac{b}{a}\sqrt[a]{{\frac{a}{b}}}}} =
\sqrt[5]{{\frac{b}{a}}}.\sqrt[{15}]{{\frac{a}{b}}} = {\left( {\frac{b}{a}}
\right)^{\frac{1}{5}}}.{\left( {\frac{a}{b}} \right)^{ - \frac{1}{{15}}}} = {\left( {\frac{b}{a}}
\right)^{\frac{1}{5}}}{\left( {\frac{b}{a}} \right)^{ - \frac{1}{{15}}}}$
$ = {\left( {\frac{b}{a}}
\right)^{\frac{1}{5} - \frac{1}{{15}}}} = {\left( {\frac{b}{a}} \right)^{\frac{2}{{15}}}}$

Bài 103868

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Bài 103868

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