Đáp án:
$\begin{array}{l}
a)Theo\,Co - si:\\
y = x + \dfrac{9}{x} \ge 2\sqrt {x.\dfrac{9}{x}} = 2\\
\Rightarrow y \ge 2\\
\Rightarrow GTNN:y = 2\\
Khi:x = \dfrac{9}{x} \Rightarrow {x^2} = 9 \Rightarrow x = 3\\
b)y = 3x + \dfrac{5}{{3x + 2}}\\
= 3x + 2 + \dfrac{5}{{3x + 2}} - 2\\
Theo\,Co - si:\\
3x + 2 + \dfrac{5}{{3x + 2}} \ge 2.\sqrt {\left( {3x + 2} \right).\dfrac{5}{{3x + 2}}} = 2\sqrt 5 \\
\Rightarrow y \ge 2\sqrt 5 - 2\\
\Rightarrow GTNn:y = 2\sqrt 5 - 2\\
Khi:3x + 2 = \dfrac{5}{{3x + 2}}\\
\Rightarrow 3x + 2 = \sqrt 5 \\
\Rightarrow x = \dfrac{{\sqrt 5 - 2}}{3}\\
e)y = \dfrac{1}{x} + \dfrac{1}{{1 - x}}\\
= \dfrac{{1 - x + x}}{{x\left( {1 - x} \right)}} = \dfrac{1}{{x\left( {1 - x} \right)}}\\
Do:x.\left( {1 - x} \right) \le \dfrac{{{{\left( {x + 1 - x} \right)}^2}}}{2}\\
x.\left( {1 - x} \right) \le \dfrac{1}{2}\\
\Rightarrow \dfrac{1}{{x\left( {1 - x} \right)}} \ge 2\\
\Rightarrow y \ge 2\\
\Rightarrow GTNN:y = 2\\
Khi:x = 1 - x\\
\Rightarrow x = \dfrac{1}{2}\\
d)y = x + \dfrac{1}{{2x - 1}}\\
= \dfrac{1}{2}.\left( {2x - 1} \right) + \dfrac{1}{2} + \dfrac{1}{{2x - 1}}\\
= \dfrac{1}{2}.\left( {2x - 1} \right) + \dfrac{1}{{2x - 1}} + \dfrac{1}{2}\\
\ge 2.\sqrt {\dfrac{1}{2}.\left( {2x - 1} \right).\dfrac{1}{{2x - 1}}} + \dfrac{1}{2}\\
\Rightarrow y \ge 2.\sqrt {\dfrac{1}{2}} + \dfrac{1}{2} = \dfrac{{2\sqrt 2 + 1}}{2}\\
\Rightarrow GTNN:y = \dfrac{{2\sqrt 2 + 1}}{2}\\
Khi:\dfrac{1}{2}\left( {2x - 1} \right) = \dfrac{1}{{2x - 1}}\\
\Rightarrow {\left( {2x - 1} \right)^2} = 2\\
\Rightarrow 2x - 1 = \sqrt 2 \\
\Rightarrow x = \dfrac{{\sqrt 2 + 1}}{2}
\end{array}$
