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Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
14,\\
a,\\
{x^2}\left( {x + 3} \right) < 0 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} \ne 0\\
x + 3 < 0
\end{array} \right. \Leftrightarrow x + 3 < 0\\
b,\\
{x^2}\left( {x + 5} \right) \ge 0 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
x + 5 \ge 0
\end{array} \right. \Leftrightarrow x + 5 \ge 0\\
c,\\
5x - 1 + \frac{1}{{x - 2}} < \frac{1}{{x - 2}} \Leftrightarrow \left\{ \begin{array}{l}
x - 2 \ne 0\\
5x - 1 < 0
\end{array} \right. \Leftrightarrow 5x - 1 < 0\\
d,\\
5x - 1 + \frac{1}{{x - 2}} > \frac{1}{{x - 2}} \Leftrightarrow \left\{ \begin{array}{l}
x - 2 \ne 0\\
5x - 1 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 2\\
5x - 1 > 0
\end{array} \right.
\end{array}\)

Vậy bất phương trình \(D\) không tương đương 

\(\begin{array}{l}
15,\\
f\left( x \right) = 5x - \frac{{x + 1}}{5} - 4 - \left( {2x - 7} \right)\\
 = 5x - \frac{1}{5}x - \frac{1}{5} - 4 - 2x + 7\\
 = \frac{{14}}{5}x - \frac{{14}}{5}\\
f\left( x \right) < 0 \Leftrightarrow \frac{{14}}{5}x - \frac{{14}}{5} < 0 \Leftrightarrow x - 1 < 0 \Leftrightarrow x < 1\\
 \Rightarrow x \in \left( { - \infty ; - 1} \right)\\
16,\\
f\left( x \right) = 5x - \frac{1}{3} - \left( {12 - \frac{{2x}}{3}} \right)\\
 = \frac{{17}}{3}x - \frac{{37}}{3}\\
f\left( x \right) > 0 \Leftrightarrow \frac{{17}}{3}x - \frac{{37}}{3} > 0 \Leftrightarrow 17x - 37 > 0 \Leftrightarrow x > \frac{{37}}{{17}}\\
x \in N;\,\,\,x < 6 \Rightarrow x \in \left\{ {3;4;5} \right\}
\end{array}\)