Đáp án:
$\begin{array}{l}
1)\dfrac{{7\left( {2\sqrt 2 - 1} \right)}}{{2\sqrt 2 + 1}} + \dfrac{{2\sqrt 2 - 2}}{{2\sqrt 2 + 2}}\\
= \dfrac{{7{{\left( {2\sqrt 2 - 1} \right)}^2}}}{{\left( {2\sqrt 2 + 1} \right)\left( {2\sqrt 2 - 1} \right)}} + \dfrac{{2\left( {\sqrt 2 - 1} \right)}}{{2\left( {\sqrt 2 + 1} \right)}}\\
= \dfrac{{7.\left( {8 - 4\sqrt 2 + 1} \right)}}{{8 - 1}} + \dfrac{{{{\left( {\sqrt 2 - 1} \right)}^2}}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}}\\
= 9 - 4\sqrt 2 + \dfrac{{2 - 2\sqrt 2 + 1}}{{2 - 1}}\\
= 9 - 4\sqrt 2 + 3 - 2\sqrt 2 \\
= 12 - 6\sqrt 2 \\
= a - b\sqrt 2 \\
\Leftrightarrow \left\{ \begin{array}{l}
a = 12\\
b = 6
\end{array} \right.\\
\Leftrightarrow a + b = 18\\
2)\sqrt[3]{{x - 5}}\\
Dkxd:x \in R
\end{array}$
