Đáp án:
$\begin{array}{l}
d)\left( {2 - \sqrt 3 } \right)\sqrt {26 + 15\sqrt 3 } - \left( {2 + \sqrt 3 } \right)\sqrt {26 - 15\sqrt 3 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {2 - \sqrt 3 } \right)\sqrt {52 + 30.\sqrt 3 } \\
- \dfrac{1}{{\sqrt 2 }}.\left( {2 + \sqrt 3 } \right)\sqrt {52 - 30\sqrt 3 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {2 - \sqrt 3 } \right)\sqrt {27 + 2.3\sqrt 3 .5 + 25} \\
- \dfrac{1}{{\sqrt 2 }}.\left( {2 + \sqrt 3 } \right)\sqrt {27 - 2.3\sqrt 3 .5 + 25} \\
= \dfrac{1}{{\sqrt 2 }}.\left( {2 - \sqrt 3 } \right).\sqrt {{{\left( {3\sqrt 3 + 5} \right)}^2}} \\
- \dfrac{1}{{\sqrt 2 }}.\left( {2 + \sqrt 3 } \right).\sqrt {{{\left( {3\sqrt 3 - 5} \right)}^2}} \\
= \dfrac{1}{{\sqrt 2 }}.\left( {2 - \sqrt 3 } \right).\left( {3\sqrt 3 + 5} \right)\\
- \dfrac{1}{{\sqrt 2 }}.\left( {2 + \sqrt 3 } \right).\left( {3\sqrt 3 - 5} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {6\sqrt 3 + 10 - 9 - 5\sqrt 3 + 6\sqrt 3 - 10 + 9 - 5\sqrt 3 } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {12\sqrt 3 - 10\sqrt 3 } \right)\\
= \dfrac{1}{{\sqrt 2 }}.2\sqrt 3 \\
= \sqrt 6 \\
e)5.{\left( {\sqrt {2 + \sqrt 3 } + \sqrt {3 - \sqrt 5 } - \sqrt {\dfrac{5}{2}} } \right)^2}\\
+ {\left( {\sqrt {2 - \sqrt 3 } + \sqrt {3 + \sqrt 5 } - \sqrt {\dfrac{3}{2}} } \right)^2}\\
= 5.\dfrac{1}{2}.{\left( {\sqrt {4 + 2\sqrt 3 } + \sqrt {6 - 2\sqrt 5 } - \sqrt 5 } \right)^2}\\
+ \dfrac{1}{2}.{\left( {\sqrt {4 - 2\sqrt 3 } + \sqrt {6 + 2\sqrt 5 } - \sqrt 3 } \right)^2}\\
= \dfrac{5}{2}.{\left( {\sqrt 3 + 1 + \sqrt 5 - 1 - \sqrt 5 } \right)^2}\\
+ \dfrac{1}{2}.{\left( {\sqrt 3 - 1 + \sqrt 5 + 1 - \sqrt 3 } \right)^2}\\
= \dfrac{5}{2}.3 + \dfrac{1}{2}.5\\
= 10
\end{array}$
