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