Đáp án:

i) \(\left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a) - 12 + 5{\left( {2x - 1} \right)^2} = 33\\
 \to 5{\left( {2x - 1} \right)^2} = 45\\
 \to {\left( {2x - 1} \right)^2} = 9\\
 \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 = 2\\
x =  - 1
\end{array} \right.\\
b) - 12 + 5{\left( {2x - 1} \right)^2} = 63\\
 \to 5{\left( {2x - 1} \right)^2} = 75\\
 \to {\left( {2x - 1} \right)^2} = 15\\
 \to \left| {2x - 1} \right| = \sqrt {15} \\
 \to \left[ \begin{array}{l}
2x - 1 = \sqrt {15} \\
2x - 1 =  - \sqrt {15} 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{{\sqrt {15}  + 1}}{2}\\
x = \dfrac{{\sqrt {15}  - 1}}{2}
\end{array} \right.\\
c) - 12 + 4{\left( {2x - 1} \right)^2}{\rm{ =  }}52\\
 \to 4{\left( {2x - 1} \right)^2}{\rm{ = 64}}\\
 \to {\left( {2x - 1} \right)^2}{\rm{ = }}16\\
 \to \left| {2x - 1} \right| = 4\\
 \to \left[ \begin{array}{l}
2x - 1 = 4\\
2x - 1 =  - 4
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
d){\left( {3x - 1} \right)^2} = 25\\
 \to \left| {3x - 1} \right| = 5\\
 \to \left[ \begin{array}{l}
3x - 1 = 5\\
3x - 1 =  - 5
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{4}{3}
\end{array} \right.\\
e){\left( {2x - 3} \right)^3} = 125\\
 \to 2x - 3 = 5\\
 \to x = 4\\
f){\left( {2x - 3} \right)^3} =  - 125\\
 \to 2x - 3 =  - 5\\
 \to x =  - 1\\
g) - 11 + 3{\left( {2x - 1} \right)^2} = 64\\
 \to 3{\left( {2x - 1} \right)^2} = 75\\
 \to {\left( {2x - 1} \right)^2} = 25\\
 \to \left| {2x - 1} \right| = 5\\
 \to \left[ \begin{array}{l}
2x - 1 = 5\\
2x - 1 =  - 5
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 3\\
x =  - 2
\end{array} \right.\\
h) - 12 + 5{\left( {2x - 1} \right)^2} = 113\\
 \to 5{\left( {2x - 1} \right)^2} = 125\\
 \to {\left( {2x - 1} \right)^2} = 25\\
 \to \left| {2x - 1} \right| = 5\\
 \to \left[ \begin{array}{l}
2x - 1 = 5\\
2x - 1 =  - 5
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 3\\
x =  - 2
\end{array} \right.\\
i) - 12 + 4{\left( {2x - 1} \right)^2} = 52\\
 \to 4{\left( {2x - 1} \right)^2} = 64\\
 \to {\left( {2x - 1} \right)^2}{\rm{ = }}16\\
 \to \left| {2x - 1} \right| = 4\\
 \to \left[ \begin{array}{l}
2x - 1 = 4\\
2x - 1 =  - 4
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x =  - \dfrac{3}{2}
\end{array} \right.
\end{array}\)