Đáp án:
$\begin{array}{l}
1)b)\\
\left( {4 - \sqrt 7 } \right)\left( {\sqrt {14} + \sqrt 2 } \right).\sqrt {4 + \sqrt 7 } \\
= \dfrac{1}{2}.\left( {8 - 2\sqrt 7 } \right).\left( {\sqrt 7 + 1} \right).\sqrt 2 .\sqrt {4 + \sqrt 7 } \\
= \dfrac{1}{2}.{\left( {\sqrt 7 - 1} \right)^2}.\left( {\sqrt 7 + 1} \right).\sqrt {8 + 2\sqrt 7 } \\
= \dfrac{1}{2}.{\left( {\sqrt 7 - 1} \right)^2}.\left( {\sqrt 7 + 1} \right).\sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} \\
= \dfrac{1}{2}.{\left( {\sqrt 7 - 1} \right)^2}.\left( {\sqrt 7 + 1} \right).\left( {\sqrt 7 + 1} \right)\\
= \dfrac{1}{2}{\left[ {\left( {\sqrt 7 - 1} \right)\left( {\sqrt 7 + 1} \right)} \right]^2}\\
= \dfrac{1}{2}.{\left( {7 - 1} \right)^2}\\
= 18\\
c)\sqrt {\dfrac{4}{{3 - \sqrt 5 }}} + \sqrt {\dfrac{4}{{3 + \sqrt 5 }}} \\
= \sqrt {\dfrac{{4\left( {3 + \sqrt 5 } \right)}}{{{3^2} - 5}}} + \sqrt {\dfrac{{4\left( {3 - \sqrt 5 } \right)}}{{{3^2} - 5}}} \\
= \sqrt {3 + \sqrt 5 } + \sqrt {3 - \sqrt 5 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {6 + 2\sqrt 5 } + \sqrt {6 - 2\sqrt 5 } } \right)\\
= \dfrac{{\sqrt 2 }}{2}.\left( {\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} } \right)\\
= \dfrac{{\sqrt 2 }}{2}.\left( {\sqrt 5 + 1 + \sqrt 5 - 1} \right)\\
= \dfrac{{\sqrt 2 }}{2}.2\sqrt 5 \\
= \sqrt {10} \\
B3)\\
b)Dkxd:x \ge \dfrac{1}{4}\\
\sqrt {4x - 1} + \sqrt {9x - \dfrac{9}{4}} = 15\\
\Leftrightarrow \sqrt {4x - 1} + \sqrt {\dfrac{{9\left( {4x - 1} \right)}}{4}} = 15\\
\Leftrightarrow \sqrt {4x - 1} + \dfrac{3}{2}.\sqrt {4x - 1} = 15\\
\Leftrightarrow \dfrac{5}{2}\sqrt {4x - 1} = 15\\
\Leftrightarrow \sqrt {4x - 1} = 6\\
\Leftrightarrow 4x - 1 = 36\\
\Leftrightarrow 4x = 37\\
\Leftrightarrow x = \dfrac{{37}}{4}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{{37}}{4}
\end{array}$
