Đáp án:
$\begin{array}{l}
4)y = 3\sqrt x + 2 \ge 2\\
\Rightarrow GTNN:y = 2 \Leftrightarrow x = 0\\
5)y = x + \sqrt x - 7\\
= x + 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{{29}}{4}\\
= {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} - \dfrac{{29}}{4}\\
Do:\sqrt x + \dfrac{1}{2} \ge \dfrac{1}{2}\\
\Rightarrow {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} \ge \dfrac{1}{4}\\
\Rightarrow y \ge \dfrac{1}{4} - \dfrac{{29}}{4} = - 7\\
\Rightarrow GTNN:y = - 7 \Leftrightarrow x = 0\\
6)y = 3\sqrt x - 2 \ge - 2\\
\Rightarrow GTNN:y = - 2 \Leftrightarrow x = 0\\
9)y = 1 - \dfrac{3}{{\sqrt x + 1}}\\
\sqrt x + 1 \ge 1\\
\Rightarrow \dfrac{1}{{\sqrt x + 1}} \le 1\\
\Rightarrow \dfrac{3}{{\sqrt x + 1}} \le 3\\
\Rightarrow - \dfrac{3}{{\sqrt x + 1}} \ge - 3\\
\Rightarrow 1 - \dfrac{3}{{\sqrt x + 1}} \ge 1 - 3 = - 2\\
\Rightarrow y \ge - 2\\
\Rightarrow GTNN:y = - 2 \Leftrightarrow x = 0\\
10)y = \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 3}}{{\sqrt x + 1}}\\
= 1 - \dfrac{3}{{\sqrt x + 1}} \ge - 2\\
\Rightarrow GTNN:y = - 2 \Leftrightarrow x = 0\\
B5:\\
7)y = \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 + 2}}{{\sqrt x + 2}}\\
= 1 + \dfrac{2}{{\sqrt x + 2}}\\
Do:\sqrt x + 2 \ge 2\\
\Rightarrow \dfrac{1}{{\sqrt x + 2}} \le \dfrac{1}{2}\\
\Rightarrow \dfrac{2}{{\sqrt x + 2}} \le 1\\
\Rightarrow 1 + \dfrac{2}{{\sqrt x + 2}} \le 2\\
\Rightarrow y \le 2\\
\Rightarrow GTLN:y = 2 \Leftrightarrow x = 0\\
9)TT:GTLN:y = 3 \Leftrightarrow x = 0\\
11)y = - 5 - \sqrt x \\
\sqrt x \ge 0\\
\Rightarrow - \sqrt x \le 0\\
\Rightarrow - 5 - \sqrt x \le - 5\\
\Rightarrow GTLN:y = - 5 \Leftrightarrow x = 0
\end{array}$
