Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
C = \left( {\frac{{\sqrt 7 .\left( {\sqrt 3 - \sqrt 1 } \right)}}{{\sqrt 3 - \sqrt 1 }} + \frac{{\sqrt 5 \left( {\sqrt 2 - 1} \right)}}{{\sqrt 2 - 1}}} \right):\frac{1}{{\sqrt 7 - \sqrt 5 }}\\
C = \left( {\sqrt 7 + \sqrt 5 } \right).\left( {\sqrt 7 + \sqrt 5 } \right)\\
C = 7 - 5\\
C = 2\\
D = \left( {3\sqrt 2 + \sqrt 6 } \right).\sqrt {6 - 3\sqrt 3 } \\
D = \left( {3 + \sqrt 3 } \right).\sqrt 2 .\sqrt {6 - 3\sqrt 3 } \\
D = \left( {3 + \sqrt 3 } \right).\sqrt {12 - 6\sqrt 3 } \\
D = \left( {3 + \sqrt 3 } \right).\sqrt {9 - 2.3\sqrt 3 + 3} \\
D = \left( {3 + \sqrt 3 } \right).\sqrt {{{\left( {3 - \sqrt 3 } \right)}^2}} \\
D = \left( {3 + \sqrt 3 } \right)\left| {3 - \sqrt 3 } \right|\\
D = \left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)\\
D = 9 - 3\\
D = 6\\
E = \sqrt {4 + \sqrt {15} } - \sqrt {4 - \sqrt {15} } - \sqrt {7 - 2\sqrt 6 } \\
\Rightarrow \sqrt 2 .E = \sqrt {8 + 2\sqrt {15} } - \sqrt {8 - 2\sqrt {15} } - \sqrt 2 .\sqrt {7 - 2\sqrt 6 } \\
\sqrt 2 E = \sqrt {{{\left( {\sqrt 3 + \sqrt 5 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - \sqrt 5 } \right)}^2}} - \sqrt 2 .\sqrt {{{\left( {\sqrt 6 - 1} \right)}^2}} \\
\sqrt 2 E = \sqrt 3 + \sqrt 5 - \left| {\sqrt 3 - \sqrt 5 } \right| - \sqrt 2 .\left| {\sqrt 6 - 1} \right|\\
\sqrt 2 E = \sqrt 3 + \sqrt 5 - \sqrt 5 + \sqrt 3 - \sqrt 2 \left( {\sqrt 6 - 1} \right)\\
\sqrt 2 E = 2\sqrt 3 - 2\sqrt 3 + \sqrt 2 = \sqrt 2 \\
\Rightarrow E = 1
\end{array}$
