Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
7,\\
\dfrac{{5 + 3\sqrt 5 }}{{3 + \sqrt 5 }} = \dfrac{{{{\sqrt 5 }^2} + 3.\sqrt 5 }}{{3 + \sqrt 5 }} = \dfrac{{\sqrt 5 \left( {\sqrt 5 + 3} \right)}}{{3 + \sqrt 5 }} = \sqrt 5 \\
8,\\
\dfrac{{3\sqrt 2 - 2\sqrt 3 }}{{\sqrt 3 - \sqrt 2 }} = \dfrac{{{{\sqrt 3 }^2}.\sqrt 2 - {{\sqrt 2 }^2}.\sqrt 3 }}{{\sqrt 3 - \sqrt 2 }} = \dfrac{{\sqrt 3 .\sqrt 2 .\left( {\sqrt 3 - \sqrt 2 } \right)}}{{\sqrt 3 - \sqrt 2 }} = \sqrt 3 .\sqrt 2 = \sqrt 6 \\
9,\\
\dfrac{{6 + 2\sqrt 6 }}{{\sqrt 3 + \sqrt 2 }} = \dfrac{{{{\sqrt 6 }^2} + 2\sqrt 6 }}{{\sqrt 3 + \sqrt 2 }} = \dfrac{{\sqrt 6 \left( {\sqrt 6 + 2} \right)}}{{\sqrt 3 + \sqrt 2 }} = \dfrac{{\sqrt 6 .\sqrt 2 .\left( {\sqrt 3 + \sqrt 2 } \right)}}{{\sqrt 3 + \sqrt 2 }} = \sqrt {12} = 2\sqrt 3 \\
10,\\
\dfrac{{6\sqrt 6 - 27}}{{2\sqrt 2 - 3\sqrt 3 }} = \dfrac{{2\sqrt 2 .3\sqrt 3 - {{\left( {3\sqrt 3 } \right)}^2}}}{{2\sqrt 2 - 3\sqrt 3 }} = \dfrac{{3\sqrt 3 \left( {2\sqrt 2 - 3\sqrt 3 } \right)}}{{2\sqrt 2 - 3\sqrt 3 }} = 3\sqrt 3
\end{array}\)
