Đáp án:
$\begin{array}{l}
B1)\\
a)\sqrt {6 - 4x - {x^2}} = x + 4\\
Dkxd:\left\{ \begin{array}{l}
6 - 4x - {x^2} \ge 0\\
x + 4 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} + 4x - 6 \le 0\\
x \ge - 4
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} \le 10\\
x \ge - 4
\end{array} \right.\\
\Rightarrow - 4 \le x \le \sqrt {10} - 2\\
Pt: \Rightarrow 6 - 4x - {x^2} = {\left( {x + 4} \right)^2}\\
\Rightarrow - {x^2} - 4x + 6 = {x^2} + 8x + 16\\
\Rightarrow 2{x^2} + 12x + 10 = 0\\
\Rightarrow {x^2} + 6x + 5 = 0\\
\Rightarrow \left( {x + 1} \right)\left( {x + 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = - 1\left( {tmdk} \right)\\
x = - 5\left( {ktm} \right)
\end{array} \right.\\
\text{Vậy}\,x = - 1\\
b)\sqrt {{x^2} + 1 - x} = 3\\
Dkxd:{x^2} + 1 - x \ge 0\left( {\text{luôn}\,\text{đúng}} \right)\\
\Rightarrow {x^2} + 1 - x = 9\\
\Rightarrow {x^2} - x - 8 = 0\\
\Rightarrow {x^2} - 2.\dfrac{1}{2}.x + \dfrac{1}{4} = 8 + \dfrac{1}{4}\\
\Rightarrow {\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{{33}}{4}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt {33} }}{4}\\
x = \dfrac{{1 - \sqrt {33} }}{4}
\end{array} \right.\\
c)\sqrt x < 4\left( {dk:x \ge 0} \right)\\
\Rightarrow x < 16\\
\text{Vậy}\,0 \le x < 16\\
d)\sqrt {x - 1} > 4\left( {dk:x \ge 1} \right)\\
\Rightarrow x - 1 > 16\\
\Rightarrow x > 17\\
\text{Vậy}\,x > 17\\
B2)\\
a)\sqrt {\dfrac{{ - 3}}{{x - 5}}} \\
\Rightarrow \dfrac{{ - 3}}{{x - 5}} \ge 0\\
\Rightarrow x - 5 < 0\\
\Rightarrow x < 5\\
b)1 - 2x \ge 0\\
\Rightarrow 2x \le 1\\
\Rightarrow x \le \dfrac{1}{2}\\
c)??\\
d)\dfrac{1}{{\sqrt {4 - x} }}\\
\Rightarrow 4 - x > 0\\
\Rightarrow x < 4
\end{array}$
