Đáp án:

$\begin{array}{l}
1)a)\sqrt {x - 2\sqrt {x - 1} }  = \sqrt {x - 1}  - 1\\
\left( {dkxd:x \ge 1} \right)\\
 \Rightarrow \sqrt {x - 1 - 2\sqrt {x - 1}  + 1}  = \sqrt {x - 1}  - 1\\
 \Rightarrow \sqrt {{{\left( {\sqrt {x - 1}  - 1} \right)}^2}}  = \sqrt {x - 1}  - 1\\
 \Rightarrow \left| {\sqrt {x - 1}  - 1} \right| = \sqrt {x - 1}  - 1\\
 \Rightarrow \sqrt {x - 1}  - 1 \ge 0\\
 \Rightarrow \sqrt {x - 1}  \ge 1\\
 \Rightarrow x \ge 2\\
b)\sqrt {4 - 2\sqrt 3 }  = x - 1\\
 \Rightarrow \sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}}  = x - 1\\
 \Rightarrow \sqrt 3  - 1 = x - 1\\
 \Rightarrow x = \sqrt 3 \\
2)a)\sqrt {25{x^2}}  = x - 12\left( {dkxd:x \ge 12} \right)\\
 \Rightarrow \left| {5x} \right| = x - 12\\
 \Rightarrow \left[ \begin{array}{l}
5x = x - 12\\
5x =  - x + 12
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x =  - 3\left( {ktm} \right)\\
x = 2\left( {ktm} \right)
\end{array} \right.\\
 \Rightarrow x \in \emptyset \\
b)\sqrt {{x^2} - 2x + 1}  = 3x + 2\left( {dk:x \ge  - \dfrac{2}{3}} \right)\\
 \Rightarrow \sqrt {{{\left( {x - 1} \right)}^2}}  = 3x + 2\\
 \Rightarrow \left| {x - 1} \right| = 3x + 2\\
 \Rightarrow \left[ \begin{array}{l}
x - 1 = 3x + 2\\
x - 1 =  - 3x - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x =  - \dfrac{3}{2}\left( {ktm} \right)\\
x =  - \dfrac{1}{4}\left( {tmdk} \right)
\end{array} \right.\\
Vay\,x =  - \dfrac{1}{4}\\
3)a)\sqrt {{{\left( { - 2x} \right)}^2}}  = 4\\
 \Rightarrow \left[ \begin{array}{l}
 - 2x = 2\\
 - 2x =  - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x =  - 1
\end{array} \right.\\
b)\sqrt {{x^2} - 2x + 1}  = \left| { - 3} \right|\\
 \Rightarrow {x^2} - 2x + 1 = 9\\
 \Rightarrow {x^2} - 2x - 8 = 0\\
 \Rightarrow \left( {x - 4} \right)\left( {x + 2} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x = 4\\
x =  - 2
\end{array} \right.\\
c)\sqrt {4{x^2} + 4x + 1}  = 2\\
 \Rightarrow \left| {2x + 1} \right| = 2\\
 \Rightarrow \left[ \begin{array}{l}
2x + 1 = 2\\
2x + 1 =  - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
d)\sqrt {3{x^2}}  = \left| { - \sqrt 6 } \right|\\
 \Rightarrow \left[ \begin{array}{l}
\sqrt 3 x = \sqrt 6 \\
\sqrt 3 x =  - \sqrt 6 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \sqrt 2 \\
x =  - \sqrt 2 
\end{array} \right.\\
4)a)4{x^2} - 4\sqrt 3 x + 3 = 0\\
 \Rightarrow {\left( {2x - \sqrt 3 } \right)^2} = 0\\
 \Rightarrow x = \dfrac{{\sqrt 3 }}{2}
\end{array}$

$\begin{array}{l}
b){x^2} + 4\sqrt 5 x + 20 = 0\\
 \Rightarrow {\left( {x + 2\sqrt 5 } \right)^2} = 0\\
 \Rightarrow x =  - 2\sqrt 5 \\
c)\sqrt {{x^4}}  = 4\\
 \Rightarrow {x^2} = 2\\
 \Rightarrow \left[ \begin{array}{l}
x = \sqrt 2 \\
x =  - \sqrt 2 
\end{array} \right.\\
5)a)\left| {2x - 1} \right| = 1 - 2x\\
 \Rightarrow 2x - 1 \le 0\\
 \Rightarrow x \le \dfrac{1}{2}\\
b)\left| {3x + 2} \right| = 3x + 2\\
 \Rightarrow 3x + 2 \ge 0\\
 \Rightarrow x \ge  - \dfrac{2}{3}\\
c)\sqrt {9{x^2} - 6x + 1}  = 3x - 1\\
 \Rightarrow \left| {3x - 1} \right| = 3x - 1\\
 \Rightarrow 3x - 1 \ge 0\\
 \Rightarrow x \ge \dfrac{1}{3}\\
d)\sqrt {4{x^2} - 12x + 9}  = 3 - 2x\\
 \Rightarrow \left| {2x - 3} \right| = 3 - 2x\\
 \Rightarrow 2x - 3 \le 0\\
 \Rightarrow x \le \dfrac{3}{2}\\
6)a)\sqrt {25{x^2}}  - 3x - 2 = 0\\
 \Rightarrow \left| {5x} \right| = 3x + 2\left( {dk:x \ge  - \dfrac{2}{3}} \right)\\
 \Rightarrow \left[ \begin{array}{l}
5x = 3x + 2\\
5x =  - 3x - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x =  - \dfrac{1}{4}\left( {tm} \right)
\end{array} \right.\\
b)\sqrt {{{\left( { - 3x} \right)}^2}}  + x - 1 = 0\\
 \Rightarrow \left| {3x} \right| = 1 - x\left( {x \le 1} \right)\\
 \Rightarrow \left[ \begin{array}{l}
3x = 1 - x\\
3x = x - 1
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\left( {tmdk} \right)\\
x =  - \dfrac{1}{2}\left( {tmdk} \right)
\end{array} \right.\\
c)\sqrt {{x^2} - 10x + 25}  = x + 4\left( {x \ge 4} \right)\\
 \Rightarrow \left| {x - 5} \right| = x + 4\\
 \Rightarrow x - 5 =  - x - 4\\
 \Rightarrow x = \dfrac{1}{2}\left( {ktm} \right)\\
d)\sqrt {{x^2} + 12x + 36}  = 2x + 5\left( {x \ge  - \dfrac{5}{2}} \right)\\
 \Rightarrow \left| {x + 6} \right| = 2x + 5\\
 \Rightarrow \left[ \begin{array}{l}
x + 6 = 2x + 5\\
x + 6 =  - 2x - 5
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = \dfrac{{ - 11}}{3}\left( {ktm} \right)
\end{array} \right.
\end{array}$