Giải thích các bước giải:
\(\begin{array}{l}
1,\\
a,\\
\frac{2}{{\sqrt 7 - 5}} + \frac{2}{{\sqrt 7 + 5}} = \frac{{2.\left( {\sqrt 7 + 5} \right) + 2.\left( {\sqrt 7 - 5} \right)}}{{\left( {\sqrt 7 - 5} \right)\left( {\sqrt 7 + 5} \right)}} = \frac{{4\sqrt 7 }}{{{{\left( {\sqrt 7 } \right)}^2} - {5^2}}} = \frac{{4\sqrt 7 }}{{ - 18}} = - \frac{{2\sqrt 7 }}{9}\\
b,\\
\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } \\
= \frac{1}{{\sqrt 2 }}.\left( {\sqrt 2 .\sqrt {2 + \sqrt 3 } + \sqrt 2 .\sqrt {2 - \sqrt 3 } } \right)\\
= \frac{1}{{\sqrt 2 }}.\left( {\sqrt {4 + 2\sqrt 3 } + \sqrt {4 - 2\sqrt 3 } } \right)\\
= \frac{1}{{\sqrt 2 }}.\left( {\sqrt {3 + 2.\sqrt 3 + 1} + \sqrt {3 - 2\sqrt 3 + 1} } \right)\\
= \frac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} } \right)\\
= \frac{1}{{\sqrt 2 }}\left( {\sqrt 3 + 1 + \sqrt 3 - 1} \right)\\
= \frac{1}{{\sqrt 2 }}.2\sqrt 3 = \sqrt 6 \\
2,\\
D = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} + \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}\\
= \frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2} + {{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}}}{{\left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}\\
= \frac{{5 + 2\sqrt {15} + 3 + 5 - 2\sqrt {15} + 3}}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}}\\
= \frac{{16}}{2}\\
= 8
\end{array}\)
