Đáp án:
$\begin{array}{l}
a)\dfrac{{\sqrt {15} - \sqrt {12} }}{{\sqrt 5 - \sqrt 2 }} - \dfrac{1}{{2 - \sqrt 3 }}\\
= \dfrac{{\sqrt 3 .\left( {\sqrt 5 - 2} \right)\left( {\sqrt 5 + \sqrt 2 } \right)}}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)}} - \dfrac{{2 + \sqrt 3 }}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}}\\
= \dfrac{{\sqrt 3 .\left( {5 + \sqrt {10} - 2\sqrt 5 - 2\sqrt 2 } \right)}}{{5 - 2}} - \dfrac{{2 + \sqrt 3 }}{{{2^2} - 3}}\\
= \dfrac{{5\sqrt 3 }}{3} + \dfrac{{\sqrt {30} }}{3} - \dfrac{{2\sqrt {15} }}{3} - \dfrac{{2\sqrt 6 }}{3} - 2 - \sqrt 3 \\
= \dfrac{{2\sqrt 3 + \sqrt {30} - 2\sqrt {15} - 2\sqrt 6 - 6}}{3}\\
a1)\dfrac{{\sqrt {15} - \sqrt {12} }}{{\sqrt 5 - 2}} - \dfrac{1}{{2 - \sqrt 3 }}\\
= \dfrac{{\sqrt 3 \left( {\sqrt 5 - \sqrt 4 } \right)}}{{\sqrt 5 - \sqrt 4 }} - \dfrac{{2 + \sqrt 3 }}{{{2^2} - 3}}\\
= \sqrt 3 - 2 - \sqrt 3 \\
= - 2\\
b)\dfrac{{1 + \sqrt 5 }}{{\sqrt {15} - \sqrt 5 + \sqrt 3 - 1}}\\
= \dfrac{{1 + \sqrt 5 }}{{\sqrt 5 \left( {\sqrt 3 - 1} \right) + \sqrt 3 - 1}}\\
= \dfrac{{1 + \sqrt 5 }}{{\left( {\sqrt 5 + 1} \right)\left( {\sqrt 3 - 1} \right)}}\\
= \dfrac{1}{{\sqrt 3 - 1}}\\
= \dfrac{{\sqrt 3 + 1}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\\
= \dfrac{{\sqrt 3 + 1}}{{3 - 1}}\\
= \dfrac{{\sqrt 3 + 1}}{2}
\end{array}$
(Có thể câu a em chép đề sai, c đã sửa đề đúng ở a1)
