Đáp án:

\(\begin{array}{l}
B2:\\
1)\left[ \begin{array}{l}
x = 5\\
x =  - 5
\end{array} \right.\\
5)\left[ \begin{array}{l}
x = 5\\
x = 1
\end{array} \right.\\
9)\left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{1}{3}
\end{array} \right.\\
2)\left[ \begin{array}{l}
x = 2\\
x =  - 2
\end{array} \right.\\
6)\left[ \begin{array}{l}
x = 6\\
x =  - 10
\end{array} \right.\\
10)\left[ \begin{array}{l}
x = 1\\
x =  - 2
\end{array} \right.\\
3)\left[ \begin{array}{l}
x = 5\\
x =  - 5
\end{array} \right.\\
7)\left[ \begin{array}{l}
x = 7\\
x =  - 5
\end{array} \right.\\
11)\left[ \begin{array}{l}
x = \sqrt 3 \\
x =  - \sqrt 3 
\end{array} \right.\\
4)\left[ \begin{array}{l}
x = 5\\
x =  - 5
\end{array} \right.\\
8)\left[ \begin{array}{l}
x = 14\\
x =  - 12
\end{array} \right.\\
B3:\\
1)x =  - 28\\
2)x = \sqrt 2 
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
B2:\\
1)\sqrt {{x^2}}  = 5\\
 \to \left| x \right| = 5\\
 \to \left[ \begin{array}{l}
x = 5\\
x =  - 5
\end{array} \right.\\
5)\sqrt {{{\left( {x - 3} \right)}^2}}  = 2\\
 \to \left| {x - 3} \right| = 2\\
 \to \left[ \begin{array}{l}
x - 3 = 2\\
x - 3 =  - 2
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 5\\
x = 1
\end{array} \right.\\
9)\sqrt {{{\left( {3x - 1} \right)}^2}}  = 2\\
 \to \left| {3x - 1} \right| = 2\\
 \to \left[ \begin{array}{l}
3x - 1 = 2\\
3x - 1 =  - 2
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{1}{3}
\end{array} \right.\\
2)\left| x \right| = 2\\
 \to \left[ \begin{array}{l}
x = 2\\
x =  - 2
\end{array} \right.\\
6)\sqrt {{{\left( {x + 2} \right)}^2}}  = 8\\
 \to \left| {x + 2} \right| = 8\\
 \to \left[ \begin{array}{l}
x + 2 = 8\\
x + 2 =  - 8
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 6\\
x =  - 10
\end{array} \right.\\
10)\sqrt {{{\left( {2x + 1} \right)}^2}}  = 3\\
 \to \left| {2x + 1} \right| = 3\\
 \to \left[ \begin{array}{l}
2x + 1 = 3\\
2x + 1 =  - 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\\
x =  - 2
\end{array} \right.\\
3)2\left| x \right| = 10\\
 \to \left| x \right| = 5\\
 \to \left[ \begin{array}{l}
x = 5\\
x =  - 5
\end{array} \right.\\
7)\sqrt {{{\left( {x - 1} \right)}^2}}  = 6\\
 \to \left| {x - 1} \right| = 6\\
 \to \left[ \begin{array}{l}
x - 1 = 6\\
x - 1 =  - 6
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 7\\
x =  - 5
\end{array} \right.\\
11)\sqrt {{x^4}}  = 3\\
 \to {x^2} = 3\\
 \to \left| x \right| = \sqrt 3 \\
 \to \left[ \begin{array}{l}
x = \sqrt 3 \\
x =  - \sqrt 3 
\end{array} \right.\\
4)3\left| x \right| = 15\\
 \to \left| x \right| = 5\\
 \to \left[ \begin{array}{l}
x = 5\\
x =  - 5
\end{array} \right.\\
8)\sqrt {{{\left( {x - 1} \right)}^2}}  = 13\\
 \to \left| {x - 1} \right| = 13\\
 \to \left[ \begin{array}{l}
x - 1 = 13\\
x - 1 =  - 13
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 14\\
x =  - 12
\end{array} \right.\\
B3:\\
1)DK:\dfrac{4}{5} \ge x\\
\sqrt {4 - 5x}  = 12\\
 \to 4 - 5x = 144\\
 \to 5x =  - 140\\
 \to x =  - 28\\
2)DK:x \ge  - \dfrac{3}{2}\\
\sqrt {2x + 3}  = 1 + \sqrt 2 \\
 \to 2x + 3 = 1 + 2\sqrt 2  + 2\\
 \to 2x = 2\sqrt 2 \\
 \to x = \sqrt 2 
\end{array}\)