Đáp án:
1) $3$
2) $4\sqrt{2}$
Giải thích các bước giải:
$\begin{array}{l}
1)\dfrac{2}{{\sqrt 3 - 1}} + \dfrac{1}{{\sqrt 3 + 2}}\\
= \dfrac{{2\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}} + \dfrac{{\sqrt 3 - 2}}{{\left( {\sqrt 3 + 2} \right)\left( {\sqrt 3 - 2} \right)}}\\
= \dfrac{{2\left( {\sqrt 3 + 1} \right)}}{2} + \dfrac{{\sqrt 3 - 2}}{{ - 1}}\\
= \sqrt 3 + 1 + 2 - \sqrt 3 \\
= 3\\
2)\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}} - \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}\\
= \dfrac{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} - \dfrac{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}\\
= \dfrac{{3 + 2\sqrt 2 }}{1} - \dfrac{{3 - 2\sqrt 2 }}{1}\\
= 4\sqrt 2
\end{array}$
