Đáp án:

\(\begin{array}{l}
21,\\
a,\,\,\,\left[ \begin{array}{l}
x =  - 2\\
x = 2
\end{array} \right.\\
b,\,\,\,\left[ \begin{array}{l}
x = 4\\
x =  - 2
\end{array} \right.\\
c,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
d,\,\,\,\,\left[ \begin{array}{l}
x = \sqrt 2 \\
x =  - \sqrt 2 
\end{array} \right.\\
22,\\
a,\,\,\,\,x = \dfrac{{\sqrt 3 }}{2}\\
b,\,\,\,x =  - 2\sqrt 5 \\
c,\,\,\,x =  \pm 2\\
d,\,\,\,x =  \pm 7\\
23,\\
a,\,\,\,\,x \le \dfrac{1}{2}\\
b,\,\,\,\,x \ge  - \dfrac{2}{3}\\
c,\,\,\,\,x \ge \dfrac{1}{3}\\
d,\,\,\,\,x \le \dfrac{3}{2}\\
24,\\
a,\,\,\,\left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{1}{4}
\end{array} \right.\\
b,\,\,\,\left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
c,\,\,\,\,x = \dfrac{1}{2}\\
d,\,\,\,\,x = 1
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
21,\\
a,\,\,\,\sqrt {{{\left( { - 2x} \right)}^2}}  = 4\\
 \Leftrightarrow \left| { - 2x} \right| = 4\\
 \Leftrightarrow \left[ \begin{array}{l}
 - 2x = 4\\
 - 2x =  - 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\\
x = 2
\end{array} \right.\\
b,\,\,\,\sqrt {{x^2} - 2x + 1}  = \left| { - 3} \right|\\
 \Leftrightarrow \sqrt {{x^2} - 2.x.1 + {1^2}}  = 3\\
 \Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2}}  = 3\\
 \Leftrightarrow \left| {x - 1} \right| = 3\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x =  - 2
\end{array} \right.\\
c,\,\,\,\,\sqrt {4{x^2} + 4x + 1}  = 2\\
 \Leftrightarrow \sqrt {{{\left( {2x} \right)}^2} + 2.2x.1 + {1^2}}  = 2\\
 \Leftrightarrow \sqrt {{{\left( {2x + 1} \right)}^2}}  = 2\\
 \Leftrightarrow \left| {2x + 1} \right| = 2\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 2\\
2x + 1 =  - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 1\\
2x =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
d,\,\,\,\,\sqrt {3{x^2}}  = \left| { - \sqrt 6 } \right|\\
 \Leftrightarrow \sqrt {{{\left( {\sqrt 3 x} \right)}^2}}  = \sqrt 6 \\
 \Leftrightarrow \left| {\sqrt 3 x} \right| = \sqrt 6 \\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt 3 x = \sqrt 6 \\
\sqrt 3 x =  - \sqrt 6 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \sqrt 2 \\
x =  - \sqrt 2 
\end{array} \right.\\
22,\\
a,\,\,\,\,4{x^2} - 4\sqrt 3 x + 3 = 0\\
 \Leftrightarrow {\left( {2x} \right)^2} - 2.2x.\sqrt 3  + {\sqrt 3 ^2} = 0\\
 \Leftrightarrow {\left( {2x - \sqrt 3 } \right)^2} = 0\\
 \Leftrightarrow 2x - \sqrt 3  = 0\\
 \Leftrightarrow x = \dfrac{{\sqrt 3 }}{2}\\
b,\,\,\,{x^2} + 4\sqrt 5 x + 20 = 0\\
 \Leftrightarrow {x^2} + 2.x.2\sqrt 5  + {\left( {2\sqrt 5 } \right)^2} = 0\\
 \Leftrightarrow {\left( {x + 2\sqrt 5 } \right)^2} = 0\\
 \Leftrightarrow x + 2\sqrt 5  = 0\\
 \Leftrightarrow x =  - 2\sqrt 5 \\
c,\,\,\,\sqrt {{x^4}}  = 4\\
 \Leftrightarrow \sqrt {{{\left( {{x^2}} \right)}^2}}  = 4\\
 \Leftrightarrow \left| {{x^2}} \right| = 4\\
 \Leftrightarrow {x^2} = 4\\
 \Leftrightarrow x =  \pm 2\\
d,\,\,\,\sqrt {\sqrt {{x^4}} }  = 7\\
 \Leftrightarrow \sqrt {\sqrt {{{\left( {{x^2}} \right)}^2}} }  = 7\\
 \Leftrightarrow \sqrt {\left| {{x^2}} \right|}  = 7\\
 \Leftrightarrow \sqrt {{x^2}}  = 7\\
 \Leftrightarrow \left| x \right| = 7\\
 \Leftrightarrow x =  \pm 7\\
23,\\
a,\,\,\,\,\left| {2x - 1} \right| = 1 - 2x\\
 \Leftrightarrow \left| {2x - 1} \right| =  - \left( {2x - 1} \right)\\
 \Leftrightarrow 2x - 1 \le 0\\
 \Leftrightarrow x \le \dfrac{1}{2}\\
b,\,\,\,\,\left| {3x + 2} \right| = 3x + 2\\
 \Leftrightarrow 3x + 2 \ge 0\\
 \Leftrightarrow x \ge  - \dfrac{2}{3}\\
c,\,\,\,\,\sqrt {9{x^2} - 6x + 1}  = 3x - 1\\
 \Leftrightarrow \sqrt {{{\left( {3x} \right)}^2} - 2.3x.1 + {1^2}}  = 3x - 1\\
 \Leftrightarrow \sqrt {{{\left( {3x - 1} \right)}^2}}  = 3x - 1\\
 \Leftrightarrow \left| {3x - 1} \right| = 3x - 1\\
 \Leftrightarrow 3x - 1 \ge 0\\
 \Leftrightarrow x \ge \dfrac{1}{3}\\
d,\,\,\,\,\sqrt {4{x^2} - 12x + 9}  = 3 - 2x\\
 \Leftrightarrow \sqrt {{{\left( {2x} \right)}^2} - 2.2x.3 + {3^2}}  = 3 - 2x\\
 \Leftrightarrow \sqrt {{{\left( {2x - 3} \right)}^2}}  = 3 - 2x\\
 \Leftrightarrow \left| {2x - 3} \right| = 3 - 2x\\
 \Leftrightarrow \left| {2x - 3} \right| =  - \left( {2x - 3} \right)\\
 \Leftrightarrow 2x - 3 \le 0\\
 \Leftrightarrow 2x \le 3\\
 \Leftrightarrow x \le \dfrac{3}{2}\\
24,\\
a,\,\,\,\sqrt {25{x^2}}  - 3x - 2 = 0\\
 \Leftrightarrow \sqrt {{{\left( {5x} \right)}^2}}  - 3x - 2 = 0\\
 \Leftrightarrow \left| {5x} \right| - 3x - 2 = 0\\
 \Leftrightarrow \left| {5x} \right| = 3x + 2\\
 \Leftrightarrow \left\{ \begin{array}{l}
3x + 2 \ge 0\\
\left[ \begin{array}{l}
5x = 3x + 2\\
5x =  - 3x - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{2}{3}\\
\left[ \begin{array}{l}
2x = 2\\
8x =  - 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{2}{3}\\
\left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{1}{4}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{1}{4}
\end{array} \right.\\
b,\,\,\,\sqrt {{{\left( { - 3x} \right)}^2}}  + x - 1 = 0\\
 \Leftrightarrow \left| { - 3x} \right| + x - 1 = 0\\
 \Leftrightarrow \left| {3x} \right| = 1 - x\\
 \Leftrightarrow \left\{ \begin{array}{l}
1 - x \ge 0\\
\left[ \begin{array}{l}
3x = 1 - x\\
3x = x - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
\left[ \begin{array}{l}
4x = 1\\
2x =  - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
\left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x =  - \dfrac{1}{2}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
c,\,\,\,\,\sqrt {{x^2} - 10x + 25}  = x + 4\\
 \Leftrightarrow \sqrt {{x^2} - 2.x.5 + {5^2}}  = x + 4\\
 \Leftrightarrow \sqrt {{{\left( {x - 5} \right)}^2}}  = x + 4\\
 \Leftrightarrow \left| {x - 5} \right| = x + 4\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 4 \ge 0\\
\left[ \begin{array}{l}
x - 5 = x + 4\\
x - 5 =  - x - 4
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 4\\
\left[ \begin{array}{l}
 - 5 = 4\,\,\,\left( L \right)\\
2x = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - 4\\
x = \dfrac{1}{2}
\end{array} \right. \Leftrightarrow x = \dfrac{1}{2}\\
d,\,\,\,\,\sqrt {{x^2} + 12x + 36}  = 2x + 5\\
 \Leftrightarrow \sqrt {{x^2} + 2.x.6 + {6^2}}  = 2x + 5\\
 \Leftrightarrow \sqrt {{{\left( {x + 6} \right)}^2}}  = 2x + 5\\
 \Leftrightarrow \left| {x + 6} \right| = 2x + 5\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x + 5 \ge 0\\
\left[ \begin{array}{l}
x + 6 = 2x + 5\\
x + 6 =  - 2x - 5
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{5}{2}\\
\left[ \begin{array}{l}
1 = x\\
3x =  - 11
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge  - \dfrac{5}{2}\\
\left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{{11}}{3}
\end{array} \right.
\end{array} \right. \Leftrightarrow x = 1
\end{array}\)