Đáp án:
$\begin{array}{l}
B2)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \# 0\\
\sqrt x - 2\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x\# 0\\
\sqrt x \# 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x\# 4
\end{array} \right.\\
Vậy\,x > 0;x\# 4\\
B3)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \# 4\\
x + 16\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x\# 16\\
x\# - 16
\end{array} \right.\\
Vậy\,x \ge 0;x\# 16\\
B4)Dkxd:\left\{ \begin{array}{l}
a \ge 0\\
\sqrt a \# 0\\
\sqrt a \# 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a \ge 0\\
a\# 0\\
a\# 4
\end{array} \right.\\
Vậy\,a > 0;a\# 4
\end{array}$
