$\begin{array}{l}
A = \dfrac{3}{{2x + 4}} - \dfrac{{x - 1}}{{{x^2} + 2x}}\\
a)\,\,DKXD:\,\,\left\{ \begin{array}{l}
2x + 4 \ne 0\\
{x^2} + 2x \ne 0
\end{array} \right. \Leftrightarrow \,\left\{ \begin{array}{l}
2\left( {x + 2} \right) \ne 0\\
x\left( {x + 2} \right) \ne 0
\end{array} \right. \Leftrightarrow \,\left\{ \begin{array}{l}
x \ne - 2\\
x \ne 0
\end{array} \right.\\
b)\,\,A = \dfrac{3}{{2x + 4}} - \dfrac{{x - 1}}{{{x^2} + 2x}}\\
= \dfrac{3}{{2\left( {x + 2} \right)}} - \dfrac{{x - 1}}{{x\left( {x + 2} \right)}}\\
= \dfrac{{3.x - \left( {x - 1} \right).2}}{{2x\left( {x + 2} \right)}} = \dfrac{{3x - 2x + 2}}{{2x\left( {x + 2} \right)}}\\
= \dfrac{{x + 2}}{{2x\left( {x + 2} \right)}} = \dfrac{1}{{2x}}\\
c)\,\,A = \dfrac{1}{4} \Leftrightarrow \dfrac{1}{{2x}} = \dfrac{1}{4} \Leftrightarrow 2x = 4 \Leftrightarrow x = 2
\end{array}$
