Đáp án:

$\begin{array}{l}
21)l)Dkxd:{x^2} - 4x - 6 \ge 0\\
 \Rightarrow \left[ \begin{array}{l}
x \ge 2 + \sqrt {10} \\
x \le 2 - \sqrt {10} 
\end{array} \right.\\
\sqrt {2{x^2} - 8x + 12}  = {x^2} - 4x - 6\\
 \Rightarrow \sqrt {2.\left( {{x^2} - 4x + 6} \right)}  = {x^2} - 4x - 6\\
 \Rightarrow \sqrt 2 .\sqrt {{x^2} - 4x + 6}  = {x^2} - 4x + 6 - 12\\
Đặt:\sqrt {{x^2} - 4x + 6}  = a\left( {a \ge 0} \right)\\
 \Rightarrow \sqrt 2 .a = {a^2} - 12\\
 \Rightarrow {a^2} - \sqrt 2 .a - 12 = 0\\
 \Rightarrow \left[ \begin{array}{l}
a = 3\sqrt 2 \left( {tm} \right)\\
a =  - 2\sqrt 2 \left( {ktm} \right)
\end{array} \right.\\
 \Rightarrow \sqrt {{x^2} - 4x + 6}  = 3\sqrt 2 \\
 \Rightarrow {x^2} - 4x + 6 = 18\\
 \Rightarrow {x^2} - 4x - 12 = 0\\
 \Rightarrow \left( {x - 6} \right)\left( {x + 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 6\left( {tm} \right)\\
x =  - 2\left( {tm} \right)
\end{array} \right.\\
Vay\,x = 6;x =  - 2\\
22)a)Dkxd:{x^2} - 3x - 4 \ge 0\\
 \Rightarrow \left( {x - 4} \right)\left( {x + 1} \right) \ge 0\\
 \Rightarrow \left[ \begin{array}{l}
x \ge 4\\
x \le  - 1
\end{array} \right.\\
\sqrt {{x^2} - 3x + 2}  = {x^2} - 3x - 4\\
 \Rightarrow \sqrt {{x^2} - 3x + 2}  = {x^2} - 3x + 2 - 6\\
Đặt:\sqrt {{x^2} - 3x + 2}  = a\left( {a \ge 0} \right)\\
 \Rightarrow a = {a^2} - 6\\
 \Rightarrow {a^2} - a - 6 = 0\\
 \Rightarrow \left( {a - 3} \right)\left( {a + 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
a = 3\left( {tm} \right)\\
a =  - 2\left( {ktm} \right)
\end{array} \right.\\
 \Rightarrow \sqrt {{x^2} - 3x + 2}  = 3\\
 \Rightarrow {x^2} - 3x + 2 = 9\\
 \Rightarrow {x^2} - 3x - 7 = 0\\
 \Rightarrow x = \dfrac{{3 \pm \sqrt {37} }}{2}\left( {tm} \right)\\
d){x^2} + x + \sqrt { - {x^2} - x - 1}  = 4\\
DKxd: - {x^2} - x - 1 \ge 0\\
 \Rightarrow {x^2} + x + 1 \le 0\\
 \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \le 0\left( {ktm} \right)
\end{array}$

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