Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
7,\\
G = {\left( {{x^2} - 4x + 3} \right)^2} + 10.\left| {{x^2} - 4x + 3} \right| + 2020\\
{\left( {{x^2} - 4x + 3} \right)^2} \ge 0,\,\,\,\forall x\\
\left| {{x^2} - 4x + 3} \right| \ge 0,\,\,\,\forall x\\
 \Rightarrow G = {\left( {{x^2} - 4x + 3} \right)^2} + 10.\left| {{x^2} - 4x + 3} \right| + 2020 \ge 0 + 10.0 + 2020 = 2020\\
 \Rightarrow {G_{\min }} = 2020 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {{x^2} - 4x + 3} \right)^2} = 0\\
\left| {{x^2} - 4x + 3} \right| = 0
\end{array} \right. \Leftrightarrow {x^2} - 4x + 3 = 0\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left( {x - 1} \right)\left( {x - 3} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.\\
8,\\
H = \left| {5 - 2x} \right|.\left( {\left| {2x - 5} \right| + 6} \right) + 88\\
 = \left| {2x - 5} \right|.\left( {\left| {2x - 5} \right| + 6} \right) + 88\\
 = {\left| {2x - 5} \right|^2} + 6.\left| {2x - 5} \right| + 88\\
{\left| {2x - 5} \right|^2} \ge 0,\,\,\,\forall x\\
\left| {2x - 5} \right| \ge 0,\,\,\,\forall x\\
 \Rightarrow H = {\left| {2x - 5} \right|^2} + 6.\left| {2x - 5} \right| + 88 \ge 0 + 6.0 + 88 = 88\\
 \Rightarrow {H_{\min }} = 88 \Leftrightarrow \left\{ \begin{array}{l}
{\left| {2x - 5} \right|^2} = 0\\
\left| {2x - 5} \right| = 0
\end{array} \right. \Leftrightarrow 2x - 5 = 0 \Leftrightarrow x = \frac{5}{2}\\
9,\\
I = {x^2} - 2x - 4\left| {x - 1} \right| + 16\\
 = \left( {{x^2} - 2x + 1} \right) - 4.\left| {x - 1} \right| + 15\\
 = {\left( {x - 1} \right)^2} - 4.\left| {x - 1} \right| + 15\\
 = {\left| {x - 1} \right|^2} - 4.\left| {x - 1} \right| + 15\\
 = \left[ {{{\left| {x - 1} \right|}^2} - 4.\left| {x - 1} \right| + 4} \right] + 11\\
 = {\left( {\left| {x - 1} \right| - 2} \right)^2} + 11 \ge 11,\,\,\,\,\forall x\\
 \Rightarrow {I_{\min }} = 11 \Leftrightarrow {\left( {\left| {x - 1} \right| - 2} \right)^2} = 0 \Leftrightarrow \left| {x - 1} \right| = 2 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 2\\
x - 1 =  - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x =  - 1
\end{array} \right.
\end{array}\)