Đáp án:
\(\begin{array}{l}
e)0\\
f)\sqrt 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
e)\sqrt {\dfrac{{4 - 2\sqrt 3 }}{{\sqrt 2 }}} - \dfrac{{\sqrt 3 }}{{\sqrt 2 }} - \dfrac{{\sqrt 3 \left( {\sqrt 3 - \sqrt 2 } \right)}}{{2\sqrt 3 - 3\sqrt 2 }}\\
= \dfrac{{\sqrt {3 - 2\sqrt 3 .1 + 1} }}{{\sqrt 2 }} - \dfrac{{\sqrt 3 }}{{\sqrt 2 }} - \dfrac{{\sqrt 3 \left( {\sqrt 3 - \sqrt 2 } \right)}}{{ - \sqrt 6 \left( {\sqrt 3 - \sqrt 2 } \right)}}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} - \sqrt 3 }}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}\\
= \dfrac{{\sqrt 3 - 1 - \sqrt 3 + 1}}{{\sqrt 2 }} = 0\\
f)\dfrac{{2\sqrt 3 }}{3} + \dfrac{{\sqrt 2 }}{3} + \dfrac{2}{{\sqrt 3 }}.\sqrt {\dfrac{{5 - 2\sqrt 6 }}{{12}}} \\
= \dfrac{{2\sqrt 3 + \sqrt 2 }}{3} + \dfrac{2}{{\sqrt 3 }}.\dfrac{{\sqrt {3 - 2.\sqrt 3 .\sqrt 2 + 2} }}{{2\sqrt 3 }}\\
= \dfrac{{2\sqrt 3 + \sqrt 2 }}{3} + \dfrac{{2\sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} }}{{2.3}}\\
= \dfrac{{2\sqrt 3 + \sqrt 2 }}{3} + \dfrac{{\sqrt 3 - \sqrt 2 }}{3}\\
= \dfrac{{3\sqrt 3 }}{3} = \sqrt 3
\end{array}\)
