Đáp án:
$\begin{array}{l}
1)a)\sqrt {2x + 7} \\
Dkxd:2x + 7 \ge 0\\
\Leftrightarrow 2x \ge - 7\\
\Leftrightarrow x \ge - \dfrac{7}{2}\\
Vậy\,x \ge \dfrac{{ - 7}}{2}\\
b)\sqrt {{x^2}.2} \\
Dkxd:2.{x^2} \ge 0\\
\Leftrightarrow {x^2} \ge 0\left( {tm} \right)\\
Vậy\,x \in R\\
c)\sqrt {\dfrac{1}{{{x^2}}}} \\
Dkxd:\dfrac{1}{{{x^2}}} \ge 0\\
\Leftrightarrow {x^2} > 0\\
\Leftrightarrow x \ne 0\\
Vậy\,x \ne 0\\
B2)\\
a)2\sqrt {{4^2}} - 5a\\
= 2.4 - 5a\\
= 8 - 5a\\
b)\sqrt {9{a^4}} + 3{a^2}\\
= 3{a^2} + 3{a^2}\\
= 6{a^2}
\end{array}$