Đáp án:
a. Min=9
b. Min=10
c. Min=1
Giải thích các bước giải:
\(\begin{array}{l}
a.A = {\left( {x - 3} \right)^2} + 9\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to {\left( {x - 3} \right)^2} + 9 \ge 9\\
\to Min = 9\\
\Leftrightarrow x - 3 = 0\\
\Leftrightarrow x = 3\\
b.B = {\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^4} + 10\\
Do:\left\{ \begin{array}{l}
{\left( {x - 1} \right)^2} \ge 0\forall x\\
{\left( {y + 2} \right)^4} \ge 0\forall y
\end{array} \right.\\
\to {\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^4} \ge 0\\
\to {\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^4} + 10 \ge 10\\
\to Min = 10\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = - 2
\end{array} \right.\\
c.C = \left| {x - 1} \right| + {\left( {2y - 1} \right)^2} + 1\\
Do:\left\{ \begin{array}{l}
\left| {x - 1} \right| \ge 0\forall x\\
{\left( {2y - 1} \right)^2} \ge 0\forall y
\end{array} \right.\\
\to \left| {x - 1} \right| + {\left( {2y - 1} \right)^2} \ge 0\\
\to \left| {x - 1} \right| + {\left( {2y - 1} \right)^2} + 1 \ge 1\\
\to Min = 1\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = \frac{1}{2}
\end{array} \right.
\end{array}\)
