Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
g,\\
\dfrac{{14}}{{\sqrt 2 + 3}} - \dfrac{2}{{1 - \sqrt 2 }}\\
= \dfrac{{14\left( {\sqrt 2 - 3} \right)}}{{\left( {\sqrt 2 + 3} \right)\left( {\sqrt 2 - 3} \right)}} - \dfrac{{2\left( {1 + \sqrt 2 } \right)}}{{\left( {1 - \sqrt 2 } \right)\left( {1 + \sqrt 2 } \right)}}\\
= \dfrac{{14\left( {\sqrt 2 - 3} \right)}}{{2 - {3^2}}} - \dfrac{{2.\left( {1 + \sqrt 2 } \right)}}{{1 - 2}}\\
= \dfrac{{14\left( {\sqrt 2 - 3} \right)}}{{ - 7}} - \dfrac{{2.\left( {1 + \sqrt 2 } \right)}}{{ - 1}}\\
= - 2.\left( {\sqrt 2 - 3} \right) + 2.\left( {1 + \sqrt 2 } \right)\\
= - 2\sqrt 2 + 6 + 2 + 2\sqrt 2 \\
= 8\\
i,\\
\dfrac{3}{{\sqrt 2 - 1}} + \dfrac{1}{{\sqrt 2 + 1}} + 1 - 2\sqrt 2 \\
= \dfrac{{3.\left( {\sqrt 2 + 1} \right) + \left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} + 1 - 2\sqrt 2 \\
= \dfrac{{3\sqrt 2 + 3 + \sqrt 2 - 1}}{{2 - 1}} + 1 - 2\sqrt 2 \\
= \dfrac{{4\sqrt 2 + 2}}{1} + 1 - 2\sqrt 2 \\
= 4\sqrt 2 + 2 + 1 - 2\sqrt 2 \\
= 3 + 2\sqrt 2
\end{array}\)
