Đáp án:
Q. \(Max = 4\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0\\
F = \dfrac{{ - 2017}}{{\sqrt x + 1}}\\
Do:x \ge 0\\
\to \sqrt x \ge 0\\
\to \sqrt x + 1 \ge 1\\
\to \dfrac{{2017}}{{\sqrt x + 1}} \le \dfrac{{2017}}{1}\\
\to \dfrac{{ - 2017}}{{\sqrt x + 1}} \ge - 2017\\
\to Min = - 2017\\
\Leftrightarrow x = 0\\
G = \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 - 1}}{{\sqrt x + 2}} = 1 - \dfrac{1}{{\sqrt x + 2}}\\
\to \sqrt x \ge 0\\
\to \sqrt x + 2 \ge 2\\
\to \dfrac{1}{{\sqrt x + 2}} \le \dfrac{1}{2}\\
\to - \dfrac{1}{{\sqrt x + 2}} \ge - \dfrac{1}{2}\\
\to 1 - \dfrac{1}{{\sqrt x + 2}} \ge \dfrac{1}{2}\\
\to Min = \dfrac{1}{2}\\
\Leftrightarrow x = 0\\
N = x - \sqrt x + 1 = x - 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do:{\left( {\sqrt x - \dfrac{1}{2}} \right)^2} \ge 0\forall x \ge 0\\
\to {\left( {\sqrt x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
\to Min = \dfrac{3}{4}\\
\Leftrightarrow \sqrt x - \dfrac{1}{2} = 0\\
\Leftrightarrow x = \dfrac{1}{4}\\
P = x + 2\sqrt x - 5\\
= x + 2\sqrt x + 1 - 6\\
= {\left( {\sqrt x + 1} \right)^2} - 6\\
Do:{\left( {\sqrt x + 1} \right)^2} \ge 0\forall x \ge 0\\
\to {\left( {\sqrt x + 1} \right)^2} - 6 \ge - 6\\
\to Min = - 6\\
\Leftrightarrow \sqrt x + 1 = 0\left( {vô lý} \right)
\end{array}\)
⇒ Hàm số không có GTLN và GTNN
\(\begin{array}{l}
Q = - x + 4\sqrt x + 3\\
= - \left( {x - 4\sqrt x - 3} \right)\\
= - \left( {x - 4\sqrt x + 1 - 4} \right)\\
= - {\left( {\sqrt x - 1} \right)^2} + 4\\
Do:{\left( {\sqrt x - 1} \right)^2} \ge 0\forall x \ge 0\\
\to - {\left( {\sqrt x - 1} \right)^2} \le 0\\
\to - {\left( {\sqrt x - 1} \right)^2} + 4 \le 4\\
\to Max = 4\\
\Leftrightarrow \sqrt x - 1 = 0\\
\Leftrightarrow x = 1
\end{array}\)
