Đáp án:
$\begin{array}{l}
1)f\left( x \right) = \dfrac{{2x - 3}}{{2{x^2}}}\\
\Rightarrow f'\left( x \right) = \dfrac{{2.2{x^2} - 4x.\left( {2x - 3} \right)}}{{2{x^2}}}\\
\Rightarrow f'\left( x \right) = \dfrac{{12x - 4{x^2}}}{{2{x^2}}} > 0\\
\Rightarrow \dfrac{{6x - 2{x^2}}}{{{x^2}}} > 0\\
\Rightarrow \dfrac{{2x\left( {3 - x} \right)}}{{{x^2}}} > 0\\
\Rightarrow 0 < x < 3\\
\Rightarrow x \in \left( {0;3} \right)\\
2)f\left( x \right) = {x^3} - {x^2} - x + 5\\
\Rightarrow f'\left( x \right) = 3{x^2} - 2x - 1\\
f'\left( x \right) < 0\\
\Rightarrow 3{x^2} - 2x - 1 < 0\\
\Rightarrow \left( {3x + 1} \right)\left( {x - 1} \right) < 0\\
\Rightarrow - \dfrac{1}{3} < x < 1\\
3)f\left( x \right) = \dfrac{1}{3}{x^3} + \dfrac{1}{2}{x^2} - 12x - 1\\
\Rightarrow f'\left( x \right) = {x^2} + x - 12 \ge 0\\
\Rightarrow \left( {x - 3} \right)\left( {x + 4} \right) \ge 0\\
\Rightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le - 4
\end{array} \right.\\
4)f\left( x \right) = {x^4} - 2x\\
\Rightarrow f'\left( x \right) = 4{x^3} - 2 = 0\\
\Rightarrow {x^3} = \dfrac{1}{2}\\
\Rightarrow x = \dfrac{1}{{\sqrt[3]{2}}} \Rightarrow 1\,nghiệm
\end{array}$
