1 câu trả lời
Đáp án: $GTNN:A = - \dfrac{9}{2}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x \ge \dfrac{1}{2}\\
A = x - 4\sqrt {2x - 1} + 3\\
= x - 2.2\sqrt 2 .\sqrt {x - \dfrac{1}{2}} + 3\\
= x - \dfrac{1}{2} - 2.\sqrt {x - \dfrac{1}{2}} .2\sqrt 2 + 8 + \dfrac{1}{2} - 8 + 3\\
= {\left( {\sqrt {x - \dfrac{1}{2}} - 2\sqrt 2 } \right)^2} - \dfrac{9}{2}\\
Do:{\left( {\sqrt {x - \dfrac{1}{2}} - 2\sqrt 2 } \right)^2} \ge 0\\
\Leftrightarrow {\left( {\sqrt {x - \dfrac{1}{2}} - 2\sqrt 2 } \right)^2} - \dfrac{9}{2} \ge - \dfrac{9}{2}\\
\Leftrightarrow A \ge - \dfrac{9}{2}\\
\Leftrightarrow GTNN:A = - \dfrac{9}{2}\,\\
Khi:\sqrt {x - \dfrac{1}{2}} = 2\sqrt 2 \\
\Leftrightarrow x - \dfrac{1}{2} = 8\\
\Leftrightarrow x = \dfrac{{17}}{2}\left( {tmdk} \right)
\end{array}$