Đáp án:
b. \(\dfrac{{\sqrt 6 + \sqrt 2 }}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}\sqrt {\dfrac{5}{{12}} - \dfrac{1}{{\sqrt 6 }}} \\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}\sqrt {\dfrac{{5 - 2\sqrt 6 }}{{12}}} \\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}\sqrt {\dfrac{{3 - 2\sqrt 3 .\sqrt 2 + 2}}{{12}}} \\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}\sqrt {\dfrac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{12}}} \\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{1}{{\sqrt 3 }}.\dfrac{{\sqrt 3 - \sqrt 2 }}{{2\sqrt 3 }}\\
= \dfrac{1}{{\sqrt 3 }} + \dfrac{1}{{3\sqrt 2 }} + \dfrac{{\sqrt 3 - \sqrt 2 }}{6}\\
= \dfrac{{2\sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 }}{6}\\
= \dfrac{{3\sqrt 3 }}{6} = \dfrac{{\sqrt 3 }}{2}\\
b.\dfrac{{2\sqrt {3 - \sqrt {3 + \sqrt {13 + 4\sqrt 3 } } } }}{{\sqrt 6 - \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 - \sqrt {3 + \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + 2.2\sqrt 3 .1 + 1} } } }}{{\sqrt 6 - \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 - \sqrt {3 + \sqrt {{{\left( {2\sqrt 3 + 1} \right)}^2}} } } }}{{\sqrt 6 - \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 - \sqrt {3 + 2\sqrt 3 + 1} } }}{{\sqrt 6 - \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} } }}{{\sqrt 6 - \sqrt 2 }}\\
= \dfrac{{2\sqrt {3 - \sqrt 3 + 1} }}{{\sqrt 6 - \sqrt 2 }}\\
= \dfrac{2}{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}} = \dfrac{{\sqrt 2 \left( {\sqrt 3 + 1} \right)}}{{3 - 1}}\\
= \dfrac{{\sqrt 6 + \sqrt 2 }}{2}
\end{array}\)
