Đáp án:
$\begin{array}{l}
1)a)\sqrt {75} - 2\sqrt {27} + \sqrt {48} \\
= 5\sqrt 3 - 2.3\sqrt 3 + 4\sqrt 3 \\
= 5\sqrt 3 - 6\sqrt 3 + 4\sqrt 3 \\
= 3\sqrt 3 \\
b)\sqrt {16b} + 3\sqrt {49b} - \sqrt {81b} \\
= 4\sqrt b + 3.7\sqrt b - 9\sqrt b \\
= 25\sqrt b - 9\sqrt b \\
= 16\sqrt b \\
c)\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} - \sqrt {18} + \dfrac{1}{2}\sqrt {32} \\
= \sqrt 2 - 1 - 3\sqrt 2 + \dfrac{1}{2}.4\sqrt 2 \\
= - 2\sqrt 2 - 1 + 2\sqrt 2 \\
= - 1\\
2)a)Dkxd:x \ge \dfrac{5}{4}\\
\Leftrightarrow \sqrt {4x - 5} = 7\\
\Leftrightarrow 4x - 5 = 49\\
\Leftrightarrow 4x = 54\\
\Leftrightarrow x = \dfrac{{27}}{2}\\
Vậy\,x = \dfrac{{27}}{2}\left( {tmdk} \right)\\
c)\sqrt {{{\left( {x + 5} \right)}^2}} = 6\\
\Leftrightarrow \left| {x + 5} \right| = 6\\
\Leftrightarrow \left[ \begin{array}{l}
x + 5 = 6\\
x + 5 = - 6
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 11
\end{array} \right.\\
Vậy\,x = 1;x = - 11
\end{array}$
