Đáp án:
$\begin{array}{l}
1)P = \dfrac{2}{{\sqrt 3 - 1}} - \sqrt {27} + \dfrac{3}{{\sqrt 3 }}\\
= \dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} - 3\sqrt 3 + \sqrt 3 \\
= \sqrt 3 + 1 - 3\sqrt 3 + \sqrt 3 \\
= 1 - \sqrt 3 \\
2)P = \dfrac{{\sqrt 5 }}{{\sqrt 5 - 2}} - 2\sqrt 5 \\
= \dfrac{{\sqrt 5 \left( {\sqrt 5 + 2} \right)}}{{5 - 4}} - 2\sqrt 5 \\
= 5 + 2\sqrt 5 - 2\sqrt 5 \\
= 5\\
3)B = \dfrac{1}{{3 + \sqrt 7 }} + \dfrac{1}{{3 - \sqrt 7 }}\\
= \dfrac{{3 - \sqrt 7 + 3 + \sqrt 7 }}{{{3^2} - 7}}\\
= \dfrac{6}{2}\\
= 3\\
4)A = \dfrac{1}{{\sqrt 6 - 2}} + \dfrac{1}{{\sqrt 6 + 2}}\\
= \dfrac{{\sqrt 6 + 2 + \sqrt 6 - 2}}{{6 - 4}}\\
= \dfrac{{2\sqrt 6 }}{2}\\
= \sqrt 6 \\
5)A = \dfrac{1}{{2 - \sqrt 3 }} + \sqrt {7 - 4\sqrt 3 } \\
= \dfrac{{2 + \sqrt 3 }}{{{2^2} - 3}} + \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} \\
= 2 + \sqrt 3 + 2 - \sqrt 3 \\
= 4\\
6)\left( {\dfrac{{\sqrt {21} - \sqrt 7 }}{{\sqrt 3 - 1}} + \dfrac{{\sqrt {10} - \sqrt 5 }}{{\sqrt 2 - 1}}} \right):\dfrac{1}{{\sqrt 7 - \sqrt 5 }}\\
= \left( {\dfrac{{\sqrt 7 \left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1}} + \dfrac{{\sqrt 5 \left( {\sqrt 2 - 1} \right)}}{{\sqrt 2 - 1}}} \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= \left( {\sqrt 7 + \sqrt 5 } \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= 7 - 5\\
= 2\\
7)P = \left( {\sqrt 3 - 1} \right).\dfrac{{3 + \sqrt 3 }}{{2\sqrt 3 }}\\
= \left( {\sqrt 3 - 1} \right).\dfrac{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{2\sqrt 3 }}\\
= \dfrac{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}{2}\\
= \dfrac{{3 - 1}}{2}\\
= 1\\
8)B = \dfrac{2}{{\sqrt 7 - \sqrt 6 }} - \sqrt {28} + \sqrt {54} \\
= \dfrac{{2\left( {\sqrt 7 + \sqrt 6 } \right)}}{{7 - 6}} - 2\sqrt 7 + 3\sqrt 6 \\
= 2\sqrt 7 + 2\sqrt 6 - 2\sqrt 7 + 3\sqrt 6 \\
= 5\sqrt 6 \\
9)A = \dfrac{4}{{\sqrt 3 - 1}} - \dfrac{2}{{\sqrt 2 + \sqrt 3 }} - \sqrt 8 \\
= \dfrac{{4\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} - \dfrac{{2\left( {\sqrt 3 - \sqrt 2 } \right)}}{{3 - 2}} - 2\sqrt 2 \\
= 2\sqrt 3 + 2 - 2\sqrt 3 + 2\sqrt 2 - 2\sqrt 2 \\
= 2\\
10)\dfrac{{50 - \sqrt {25} }}{{\sqrt {36} }} = \dfrac{{50 - 5}}{6} = \dfrac{{15}}{2}
\end{array}$