Giải bất phương trình: \(\frac{{3x - 1}}{{x + 3}} > 2\).

Ta có \(\frac{{3x + 17}}{{10}} > \frac{{5x + 22}}{{15}}\)

\(\begin{array}{l}\frac{{3\left( {3x + 17} \right)}}{{30}} > \frac{{2\left( {5x + 22} \right)}}{{30}}\\9x + 51 > 10x + 44\\9x - 10x > 44 - 51\end{array}\)

\(\begin{array}{l} - x >  - 7\\x < 7\left( {\rm{*}} \right)\end{array}\)

Ta có  \(\frac{{x - 4}}{{30}} - 1 > \frac{{2x - 27}}{{24}}\)

\(\begin{array}{l}\frac{{4\left( {x - 4} \right)}}{{120}} - \frac{{120}}{{120}} > \frac{{5\left( {2x - 27} \right)}}{{120}}\\4x - 16 - 120 > 10x - 135\end{array}\)

\(\begin{array}{l}4x - 10x > 16 + 120 - 135\\ - 6x > 1\end{array}\)

\(x <  - \frac{1}{6}\left( {{\rm{**}}} \right)\)

a) \(\frac{{5x + 2}}{5} < \frac{{4x - 3}}{4}\)

\(\begin{array}{l}\frac{{4\left( {5x + 2} \right)}}{{20}} < \frac{{5\left( {4x - 3} \right)}}{{20}}\\20x + 8 < 20x - 15\\20x - 20x <  - 8 - 15\\0x <  - 23\end{array}\)

Vậy bất phương trình này vô nghiệm.

b) \(\frac{{3\left( {2x + 1} \right)}}{{20}} + 1 < \frac{{3x + 13}}{{10}}\)

\[\begin{array}{l}\frac{{3\left( {2x + 1} \right)}}{{20}} + \frac{{20}}{{20}} < \frac{{2\left( {3x + 13} \right)}}{{20}}\\6x + 3 + 20 < 6x + 26\\6x - 6x < 26 - 3 - 20\\0x < 3\end{array}\]

Vậy bất phương trình này có nghiệm bất kì.