Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
c,\,\,\,\,\,\left( {x \ne 0;\,\,x \ne - 2} \right)\\
\frac{2}{{x + 2}} + \frac{1}{2} \le \frac{{ - 4}}{{{x^2} + 2x}}\\
\Leftrightarrow \frac{2}{{x + 2}} + \frac{1}{2} + \frac{4}{{x\left( {x + 2} \right)}} \le 0\\
\Leftrightarrow \frac{{2.2x + x\left( {x + 2} \right) + 4.2}}{{2x\left( {x + 2} \right)}} \le 0\\
\Leftrightarrow \frac{{4x + {x^2} + 2x + 8}}{{2x\left( {x + 2} \right)}} \le 0\\
\Leftrightarrow \frac{{{x^2} + 6x + 8}}{{x\left( {x + 2} \right)}} \le 0\\
\Leftrightarrow \frac{{\left( {x + 2} \right)\left( {x + 4} \right)}}{{x\left( {x + 2} \right)}} \le 0\\
\Leftrightarrow \frac{{x + 4}}{x} \le 0\\
\Leftrightarrow - 4 \le x < 0\\
g,\,\,\,\,\,\,\left( {x \ne 6;\,\,x \ne - 5} \right)\\
\frac{{{x^4} - 3{x^3} + 2{x^2}}}{{{x^2} - x - 30}} > 0\\
\Leftrightarrow \frac{{{x^2}\left( {{x^2} - 3x + 2} \right)}}{{\left( {x - 6} \right)\left( {x + 5} \right)}} > 0\\
\Leftrightarrow \frac{{{x^2}\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 6} \right)\left( {x + 5} \right)}} > 0\\
{x^2} \ge 0 \Rightarrow \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 6} \right)\left( {x + 5} \right)}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 6\\
1 < x < 2\\
x < - 5
\end{array} \right.
\end{array}\)
