Đáp án:
$A = \dfrac{{6 + 3\sqrt 3 }}{{1 + \sqrt 3 }}$
Giải thích các bước giải:
Ta có:
$\sqrt {\dfrac{2}{{2 + \sqrt 3 }}} = \sqrt {\dfrac{4}{{4 + 2\sqrt 3 }}} = \sqrt {\dfrac{4}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}} = \dfrac{2}{{1 + \sqrt 3 }}$
Khi đó:
$\begin{array}{l}
A = \dfrac{{{{\left( {\sqrt {\dfrac{2}{{2 + \sqrt 3 }}} + 1} \right)}^2}}}{{\sqrt {\dfrac{2}{{2 + \sqrt 3 }}} }}\\
= \dfrac{{{{\left( {\dfrac{2}{{1 + \sqrt 3 }} + 1} \right)}^2}}}{{\dfrac{2}{{1 + \sqrt 3 }}}}\\
= \dfrac{{{{\left( {\dfrac{{3 + \sqrt 3 }}{{1 + \sqrt 3 }}} \right)}^2}}}{{\dfrac{2}{{1 + \sqrt 3 }}}}\\
= {\left( {\dfrac{{3 + \sqrt 3 }}{{1 + \sqrt 3 }}} \right)^2}.\dfrac{{1 + \sqrt 3 }}{2}\\
= \dfrac{{{{\left( {3 + \sqrt 3 } \right)}^2}}}{{2\left( {1 + \sqrt 3 } \right)}}\\
= \dfrac{{12 + 6\sqrt 3 }}{{2\left( {1 + \sqrt 3 } \right)}}\\
= \dfrac{{6 + 3\sqrt 3 }}{{1 + \sqrt 3 }}
\end{array}$
Vậy $A = \dfrac{{6 + 3\sqrt 3 }}{{1 + \sqrt 3 }}$
