Đáp án:
$\begin{array}{l}
1)\dfrac{{\sqrt {21} + 3}}{{\sqrt 7 + \sqrt 3 }} - \dfrac{{21}}{{\sqrt 7 }} + \dfrac{8}{{\sqrt 7 - \sqrt 3 }}\\
= \dfrac{{\sqrt 3 \left( {\sqrt 7 + \sqrt 3 } \right)}}{{\sqrt 7 + \sqrt 3 }} - 3\sqrt 7 + \dfrac{{8\left( {\sqrt 7 + \sqrt 3 } \right)}}{{7 - 3}}\\
= \sqrt 3 - 3\sqrt 7 + 2\left( {\sqrt 7 + \sqrt 3 } \right)\\
= 3\sqrt 3 - \sqrt 7 \\
2)\left( {3 + \dfrac{{2 - \sqrt 2 }}{{1 - \sqrt 2 }}} \right)\left( {\dfrac{{\sqrt 6 + \sqrt {10} }}{{\sqrt 5 + \sqrt 3 }} + 3} \right)\\
= \left( {3 + \dfrac{{\sqrt 2 \left( {\sqrt 2 - 1} \right)}}{{1 - \sqrt 2 }}} \right)\left( {\dfrac{{\sqrt 2 \left( {\sqrt 3 + \sqrt 5 } \right)}}{{\sqrt 5 + \sqrt 3 }} + 3} \right)\\
= \left( {3 - \sqrt 2 } \right)\left( {\sqrt 2 + 3} \right)\\
= {3^2} - 2\\
= 7\\
3)\dfrac{2}{{\sqrt 3 - 1}} - \sqrt {\dfrac{2}{{2 + \sqrt 3 }}} \\
= \dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} - \sqrt {\dfrac{4}{{4 + 2\sqrt 3 }}} \\
= \sqrt 3 + 1 - \dfrac{2}{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}\\
= \sqrt 3 + 1 - \dfrac{2}{{\sqrt 3 + 1}}\\
= \sqrt 3 + 1 - \dfrac{{2\left( {\sqrt 3 - 1} \right)}}{{3 - 1}}\\
= \sqrt 3 + 1 - \left( {\sqrt 3 - 1} \right)\\
= 2\\
4)\sqrt {\dfrac{2}{{3\sqrt 5 + 7}}} + \dfrac{2}{{3 - \sqrt 5 }}\\
= \sqrt {\dfrac{4}{{6\sqrt 5 + 14}}} + \dfrac{{2\left( {3 + \sqrt 5 } \right)}}{{{3^2} - 5}}\\
= \dfrac{2}{{\sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}} }} + \dfrac{{3 + \sqrt 5 }}{2}\\
= \dfrac{2}{{3 + \sqrt 5 }} + \dfrac{{3 + \sqrt 5 }}{2}\\
= \dfrac{{2\left( {3 - \sqrt 5 } \right)}}{{{3^2} - 5}} + \dfrac{{3 + \sqrt 5 }}{2}\\
= \dfrac{{3 - \sqrt 5 }}{2} + \dfrac{{3 + \sqrt 5 }}{2}\\
= 3\\
5)\dfrac{1}{{\sqrt 5 - \sqrt 3 }} + \dfrac{{5\sqrt 3 - 3\sqrt 5 }}{{2\sqrt {15} }} - \dfrac{{10}}{{\sqrt 5 }}\\
= \dfrac{{\sqrt 5 + \sqrt 3 }}{{5 - 3}} + \dfrac{{\sqrt {15} \left( {\sqrt 5 - \sqrt 3 } \right)}}{{2\sqrt {15} }} - 2\sqrt 5 \\
= \dfrac{{\sqrt 5 + \sqrt 3 }}{2} + \dfrac{{\sqrt 5 - \sqrt 3 }}{2} - 2\sqrt 5 \\
= \sqrt 5 - 2\sqrt 5 \\
= - \sqrt 5
\end{array}$
