\(\begin{array}{l} B = \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} = \sqrt x + 1 + \dfrac{1}{{\sqrt x }}\\ = \sqrt x + \dfrac{1}{{\sqrt x }} + 1\\ Do:x > 0\\ \to Co – si:\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} \\ \to \sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\\ \to \sqrt x + \dfrac{1}{{\sqrt x }} + 1 \ge 3\\ \to MinB = 3\\ \Leftrightarrow \sqrt x = \dfrac{1}{{\sqrt x }}\\ \Leftrightarrow x = 1\\ C = \dfrac{{3\left( {\sqrt x + 1} \right) – 1}}{{\sqrt x + 1}} = 3 – \dfrac{1}{{\sqrt x + 1}}\\ MinC \Leftrightarrow \left( {\dfrac{1}{{\sqrt x + 1}}} \right)\max \\ \Leftrightarrow \left( {\sqrt x + 1} \right)\min \\ \Leftrightarrow \sqrt x + 1 = 1\\ \Leftrightarrow x = 0\left( l \right) \end{array}\)
Mít Mít
Đáp án:
\(MinB = 3\)
Giải thích các bước giải:
\(\begin{array}{l}
B = \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} = \sqrt x + 1 + \dfrac{1}{{\sqrt x }}\\
= \sqrt x + \dfrac{1}{{\sqrt x }} + 1\\
Do:x > 0\\
\to Co – si:\sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{1}{{\sqrt x }}} \\
\to \sqrt x + \dfrac{1}{{\sqrt x }} \ge 2\\
\to \sqrt x + \dfrac{1}{{\sqrt x }} + 1 \ge 3\\
\to MinB = 3\\
\Leftrightarrow \sqrt x = \dfrac{1}{{\sqrt x }}\\
\Leftrightarrow x = 1\\
C = \dfrac{{3\left( {\sqrt x + 1} \right) – 1}}{{\sqrt x + 1}} = 3 – \dfrac{1}{{\sqrt x + 1}}\\
MinC \Leftrightarrow \left( {\dfrac{1}{{\sqrt x + 1}}} \right)\max \\
\Leftrightarrow \left( {\sqrt x + 1} \right)\min \\
\Leftrightarrow \sqrt x + 1 = 1\\
\Leftrightarrow x = 0\left( l \right)
\end{array}\)
⇒ Không tồn tại giá trị x để C đạt GTNN