Đáp án:
\(Min = 4 + \sqrt 2 \)
Giải thích các bước giải:
\(\begin{array}{l}
A = \sqrt {2{x^2} - 4x + 18} + \sqrt 2 \\
= \sqrt {2\left( {{x^2} - 2x + 9} \right)} + \sqrt 2 \\
= \sqrt {2\left( {{x^2} - 2x + 1 + 8} \right)} + \sqrt 2 \\
= \sqrt {2{{\left( {x - 1} \right)}^2} + 16} + \sqrt 2 \\
Do:{\left( {x - 1} \right)^2} \ge 0\forall x\\
\to 2{\left( {x - 1} \right)^2} \ge 0\\
\to 2{\left( {x - 1} \right)^2} + 16 \ge 16\\
\to \sqrt {2{{\left( {x - 1} \right)}^2} + 16} \ge \sqrt {16} \\
\to \sqrt {2{{\left( {x - 1} \right)}^2} + 16} \ge 4\\
\to \sqrt {2{{\left( {x - 1} \right)}^2} + 16} + \sqrt 2 \ge 4 + \sqrt 2 \\
\to Min = 4 + \sqrt 2 \\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1
\end{array}\)
