Đáp án:
B5:
c) Min=2
Giải thích các bước giải:
\(\begin{array}{l}
B5:\\
a)A = {x^2} - 6x + 9 + 2\\
= {\left( {x - 3} \right)^2} + 2\\
Do:{\left( {x - 3} \right)^2} \ge 0\forall x\\
\to {\left( {x - 3} \right)^2} + 2 \ge 2\\
\to Min = 2\\
\Leftrightarrow x = 3\\
b)B = {x^2} - 20x + 100 + 1\\
= {\left( {x - 10} \right)^2} + 1\\
Do:{\left( {x - 10} \right)^2} \ge 0\forall x\\
\to {\left( {x - 10} \right)^2} + 1 \ge 1\\
\to Min = 1\\
\Leftrightarrow x = 10\\
c)C = {x^2} + {\left( { - 2y} \right)^2} + 25 - 4xy + 2.x.5 + 2.\left( { - 2y} \right).5 + {y^2} - 2y + 1 + 2\\
= {\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} + 2\\
Do:{\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} \ge 0\forall x;y\\
\to {\left( {x - 2y + 5} \right)^2} + {\left( {y - 1} \right)^2} + 2 \ge 2\\
\to Min = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 2y + 5 = 0\\
y - 1 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 1\\
x = - 3
\end{array} \right.
\end{array}\)
