Đáp án:

$\begin{array}{l}
4)a)\sqrt {3 - 2\sqrt 2 }  = \sqrt {2 - 2\sqrt 2  + 1} \\
 = \sqrt {{{\left( {\sqrt 2  - 1} \right)}^2}} \\
 = \sqrt 2  - 1\\
b)\sqrt {3 + 2\sqrt 2 }  = \sqrt {{{\left( {\sqrt 2  + 1} \right)}^2}}  = \sqrt 2  + 1\\
c)\sqrt {6 + 2\sqrt 5 }  = \sqrt {{{\left( {\sqrt 5  + 1} \right)}^2}}  = \sqrt 5  + 1\\
d)\sqrt {8 - 2\sqrt 7 }  = \sqrt {{{\left( {\sqrt 7  - 1} \right)}^2}}  = \sqrt 7  - 1\\
e)\sqrt {7 + 4\sqrt 3 }  = \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}}  = 2 + \sqrt 3 \\
f)\sqrt {11 + 6\sqrt 2 }  = \sqrt {9 + 2.3.\sqrt 2  + 2} \\
 = \sqrt {{{\left( {3 + \sqrt 2 } \right)}^2}}  = 3 + \sqrt 2 \\
g)\sqrt {6 - 4\sqrt 2 }  = \sqrt {4 - 2.2\sqrt 2  + 2}  = \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}}  = 2 - \sqrt 2 \\
h)\sqrt {14 - 6\sqrt 5 }  = \sqrt {9 - 2.3.\sqrt 5  + 5}  = \sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}}  = 3 - \sqrt 5 \\
i)\sqrt {9 - 4\sqrt 5 }  = \sqrt {{{\left( {\sqrt 5  - 2} \right)}^2}}  = \sqrt 5  - 2\\
5)\\
a)\sqrt {{{\left( {x - 1} \right)}^2}}  = 5\\
 \Leftrightarrow \left| {x - 1} \right| = 5\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 5\\
x - 1 =  - 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x =  - 4
\end{array} \right.\\
Vậy\,x = 6;x =  - 4\\
b)\sqrt {4{x^2} - 12x + 9}  = 4\\
 \Leftrightarrow \sqrt {{{\left( {2x - 3} \right)}^2}}  = 4\\
 \Leftrightarrow \left| {2x - 3} \right| = 4\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 4\\
2x - 3 =  - 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 1}}{2};x = \dfrac{7}{2}\\
c)\sqrt {{x^2} + 6x + 9}  = 3x - 6\left( {dk:x \ge 2} \right)\\
 \Leftrightarrow \sqrt {{{\left( {x + 3} \right)}^2}}  = 3x - 6\\
 \Leftrightarrow \left| {x + 3} \right| = 3x - 6\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 3x - 6\\
x + 3 =  - 3x + 6
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{9}{2}\\
x = \dfrac{3}{4}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{9}{2}\\
e)Dk:x \ge 3\\
\sqrt {10\left( {x - 3} \right)}  = 20\\
 \Leftrightarrow 10\left( {x - 3} \right) = 400\\
 \Leftrightarrow x - 3 = 40\\
 \Leftrightarrow x = 43\left( {tmdk} \right)\\
Vậy\,x = 43\\
d)\sqrt {{x^2} - 4x + 4}  = 2x - 5\left( {dk:x \ge \dfrac{5}{2}} \right)\\
 \Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}}  = 2x - 5\\
 \Leftrightarrow \left| {x - 2} \right| = 2x - 5\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 2x - 5\\
x - 2 =  - 2x + 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = \dfrac{7}{3}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 3
\end{array}$