Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^2} = {\left( { - x} \right)^2} = {\left| x \right|^2}\\
1,\\
A = {\left( {x - 2} \right)^2} - 6.\left| {x - 2} \right| + 15\\
= {\left| {x - 2} \right|^2} - 6.\left| {x - 2} \right| + 15\\
= \left( {{{\left| {x - 2} \right|}^2} - 6.\left| {x - 2} \right| + 9} \right) + 6\\
= {\left( {\left| {x - 2} \right| - 3} \right)^2} + 6 \ge 6,\,\,\,\forall x\\
\Rightarrow {A_{\min }} = 6 \Leftrightarrow {\left( {\left| {x - 2} \right| - 3} \right)^2} = 0 \Leftrightarrow \left| {x - 2} \right| = 3 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 = - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = - 1
\end{array} \right.\\
2,\\
B = {\left( {5 - x} \right)^2} - 4.\left| {x - 5} \right| + 7\\
= {\left( {x - 5} \right)^2} - 4.\left| {x - 5} \right| + 7\\
= {\left| {x - 5} \right|^2} - 4.\left| {x - 5} \right| + 7\\
= \left( {{{\left| {x - 5} \right|}^2} - 4.\left| {x - 5} \right| + 4} \right) + 3\\
= {\left( {\left| {x - 5} \right| - 2} \right)^2} + 3 \ge 3,\,\,\,\forall x\\
\Rightarrow {B_{\min }} = 3 \Leftrightarrow {\left( {\left| {x - 5} \right| - 2} \right)^2} = 0 \Leftrightarrow \left| {x - 5} \right| = 2 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 2\\
x - 5 = - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x = 3
\end{array} \right.\\
3,\\
C = {\left( {2x - 1} \right)^2} - \left| {6x - 3} \right| + 20\\
= {\left| {2x - 1} \right|^2} - \left| {3.\left( {2x - 1} \right)} \right| + 20\\
= {\left| {2x - 1} \right|^2} - 3.\left| {2x - 1} \right| + 20\\
= \left( {{{\left| {2x - 1} \right|}^2} - 3.\left| {2x - 1} \right| + \dfrac{9}{4}} \right) + \dfrac{{71}}{4}\\
= {\left( {\left| {2x - 1} \right| - \dfrac{3}{2}} \right)^2} + \dfrac{{71}}{4} \ge \dfrac{{71}}{4},\,\,\,\,\forall x\\
\Rightarrow {C_{\min }} = \dfrac{{71}}{4} \Leftrightarrow {\left( {\left| {2x - 1} \right| - \dfrac{3}{2}} \right)^2} = 0 \Leftrightarrow \left| {2x - 1} \right| = \dfrac{3}{2} \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = \dfrac{3}{2}\\
2x - 1 = - \dfrac{3}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{4}\\
x = - \dfrac{1}{4}
\end{array} \right.
\end{array}\)
