Đáp án:
$\begin{array}{l}
b)x > \frac{3}{2} \Rightarrow 3x - 2 > 0\\
y = x + \frac{4}{{3x - 2}} = \frac{1}{3}.\left( {3x - 2} \right) + \frac{2}{3} + \frac{4}{{3x - 2}}\\
= \frac{1}{3}.\left( {3x - 2} \right) + \frac{4}{{3x - 2}} + \frac{2}{3}\\
Theo\,Cô - si:\frac{1}{3}\left( {3x - 2} \right) + \frac{4}{{3x - 2}} \ge 2.\sqrt {\frac{1}{3}\left( {3x - 2} \right).\frac{4}{{3x - 2}}} = 2.\frac{2}{{\sqrt 3 }} = \frac{4}{{\sqrt 3 }}\\
\Rightarrow y \ge \frac{4}{{\sqrt 3 }} + \frac{2}{3} = \frac{{4\sqrt 3 + 2}}{3}\\
\Rightarrow GTNN:y = \frac{{4\sqrt 3 + 2}}{3}\\
c)y = 5x + \frac{2}{{2x - 3}}\\
= \frac{5}{2}\left( {2x - 3} \right) + \frac{{15}}{2} + \frac{2}{{2x - 3}}\\
= \frac{5}{2}\left( {2x - 3} \right) + \frac{2}{{2x - 3}} + \frac{{15}}{2}\\
\ge 2.\sqrt 5 + \frac{{15}}{2}\\
\Rightarrow GTNN:y = 2\sqrt 5 + \frac{{15}}{2}
\end{array}$
