Đáp án:
c) \(MinC = 2\)
Giải thích các bước giải:
\(\begin{array}{l}
B5:\\
a)Do:{\left( {x - 2} \right)^2} \ge 0\forall x\\
\to {\left( {x - 2} \right)^2} - 1 \ge - 1\\
\to MinA = - 1\\
\Leftrightarrow x - 2 = 0\\
\to x = 2\\
b)Do:\left\{ \begin{array}{l}
{\left( {x + 3} \right)^2} \ge 0\forall x\\
{\left( {y - 5} \right)^4} \ge 0\forall y
\end{array} \right.\\
\to {\left( {x + 3} \right)^2} + {\left( {y - 5} \right)^4} \ge 0\\
\to {\left( {x + 3} \right)^2} + {\left( {y - 5} \right)^4} + 6 \ge 6\\
\to MinB = 6\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 3\\
y = 5
\end{array} \right.\\
c)Do:\left\{ \begin{array}{l}
{\left( {{x^2} - 9} \right)^2} \ge 0\forall x\\
\left| {y - 2} \right| \ge 0\forall y
\end{array} \right.\\
\to {\left( {{x^2} - 9} \right)^2} + \left| {y - 2} \right| \ge 0\\
\to {\left( {{x^2} - 9} \right)^2} + \left| {y - 2} \right| + 2 \ge 2\\
\to MinC = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 9 = 0\\
y - 2 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 3\\
x = - 3
\end{array} \right.\\
y = 2
\end{array} \right.
\end{array}\)
