Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
{x^2} = {\left( { - x} \right)^2} = {\left| x \right|^2}\\
4,\\
D = {\left( {3x - 1} \right)^2} - 2.\left| {3x - 1} \right| + 12\\
 = {\left| {3x - 1} \right|^2} - 2.\left| {3x - 1} \right| + 12\\
 = \left( {{{\left| {3x - 1} \right|}^2} - 2.\left| {3x - 1} \right| + 1} \right) + 11\\
 = {\left( {\left| {3x - 1} \right| - 1} \right)^2} + 11 \ge 11,\,\,\,\forall x\\
 \Rightarrow {D_{\min }} = 11 \Leftrightarrow {\left( {\left| {3x - 1} \right| - 1} \right)^2} = 0 \Leftrightarrow \left| {3x - 1} \right| = 1 \Leftrightarrow \left[ \begin{array}{l}
3x - 1 = 1\\
3x - 1 =  - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{2}{3}\\
x = 0
\end{array} \right.\\
5,\\
E = {\left( {3x - 1} \right)^2} - 5.\left| {6x - 2} \right| - 25\\
 = {\left( {3x - 1} \right)^2} - 5.\left| {2.\left( {3x - 1} \right)} \right| - 25\\
 = {\left| {3x - 1} \right|^2} - 10.\left| {3x - 1} \right| - 25\\
 = \left( {{{\left| {3x - 1} \right|}^2} - 10.\left| {3x - 1} \right| + 25} \right) - 50\\
 = {\left( {\left| {3x - 1} \right| - 5} \right)^2} - 50 \ge  - 50,\,\,\,\,\forall x\\
 \Rightarrow {E_{\min }} =  - 50 \Leftrightarrow {\left( {\left| {3x - 1} \right| - 5} \right)^2} = 0 \Leftrightarrow \left| {3x - 1} \right| = 5 \Leftrightarrow \left[ \begin{array}{l}
3x - 1 = 5\\
3x - 1 =  - 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - \frac{4}{3}
\end{array} \right.\\
6,\\
F = {\left( {x - 3} \right)^2} + 2.\left| {3 - x} \right| + 6\\
{\left( {x - 3} \right)^2} \ge 0,\,\,\,\forall x\\
\left| {3 - x} \right| \ge 0,\,\,\,\forall x\\
 \Rightarrow F = {\left( {x - 3} \right)^2} + 2.\left| {3 - x} \right| + 6 \ge {0^2} + 2.0 + 6 = 6\\
 \Rightarrow {F_{\min }} = 6 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 3} \right)^2} = 0\\
\left| {3 - x} \right| = 0
\end{array} \right. \Leftrightarrow x = 3
\end{array}\)