Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
A = \left| {x + y - 3} \right| + {\left( {x - 1} \right)^2} - 11\\
\left| {x + y - 3} \right| \ge 0,\,\,\,\forall x,y\\
{\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
\Rightarrow \left| {x + y - 3} \right| + {\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x,y\\
\Rightarrow \left| {x + y - 3} \right| + {\left( {x - 1} \right)^2} - 11 \ge - 11,\,\,\,\forall x,y\\
\Rightarrow A \ge - 11,\,\,\,\forall x,y\\
\Rightarrow {A_{\min }} = - 11 \Leftrightarrow \left\{ \begin{array}{l}
\left| {x + y - 3} \right| = 0\\
{\left( {x - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x + y - 3 = 0\\
x - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 2
\end{array} \right.\\
2,\\
a,\\
A = 5 - \left| {2x - 1} \right|\\
\left| {2x - 1} \right| \ge 0,\,\,\forall x \Rightarrow 5 - \left| {2x - 1} \right| \le 5,\,\,\,\forall x\\
\Rightarrow {A_{\max }} = 5 \Leftrightarrow \left| {2x - 1} \right| = 0 \Leftrightarrow x = \dfrac{1}{2}\\
b,\\
B = \dfrac{1}{{\left| {x - 2} \right| + 3}}\\
\left| {x - 2} \right| \ge 0,\,\,\forall x \Rightarrow \left| {x - 2} \right| + 3 \ge 3,\,\,\forall x\\
\Rightarrow \dfrac{1}{{\left| {x - 2} \right| + 3}} \le \dfrac{1}{3},\,\,\,\forall x\\
\Rightarrow B \le \dfrac{1}{3},\,\,\,\forall x\\
\Rightarrow {B_{\max }} = \dfrac{1}{3} \Leftrightarrow \left| {x - 2} \right| = 0 \Leftrightarrow x = 2
\end{array}\)
