Giải thích các bước giải:
$\begin{array}{l}
d)\dfrac{1}{{3x - 2}} - \dfrac{{3x - 6}}{{4 - 9{x^2}}}\left( {DK:x \ne \pm \dfrac{2}{3}} \right)\\
= \dfrac{1}{{3x - 2}} + \dfrac{{3x - 6}}{{9{x^2} - 4}}\\
= \dfrac{1}{{3x - 2}} + \dfrac{{3x - 6}}{{\left( {3x - 2} \right)\left( {3x + 2} \right)}}\\
= \dfrac{{3x + 2 + 3x - 6}}{{\left( {3x - 2} \right)\left( {3x + 2} \right)}}\\
= \dfrac{{6x - 4}}{{\left( {3x - 2} \right)\left( {3x + 2} \right)}}\\
= \dfrac{{2\left( {3x - 2} \right)}}{{\left( {3x - 2} \right)\left( {3x + 2} \right)}}\\
= \dfrac{2}{{3x + 2}}\\
e)\dfrac{{x + 3}}{{{x^2} - 1}} - \dfrac{{x + 1}}{{{x^2} - x}}\left( {DK:x \ne \left\{ {0; \pm 1} \right\}} \right)\\
= \dfrac{{x + 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{{x + 1}}{{x\left( {x - 1} \right)}}\\
= \dfrac{{x\left( {x + 3} \right) - \left( {x + 1} \right)\left( {x + 1} \right)}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + 3x - {x^2} - 2x - 1}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{x - 1}}{{x\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{1}{{x\left( {x + 1} \right)}}\\
f)\dfrac{3}{{2x + 6}} - \dfrac{{x - 6}}{{2{x^2} + 6x}}\left( {DK:x \ne \left\{ { - 3;0} \right\}} \right)\\
= \dfrac{3}{{2x + 6}} - \dfrac{{x - 6}}{{x\left( {2x + 6} \right)}}\\
= \dfrac{{3x - \left( {x - 6} \right)}}{{x\left( {2x + 6} \right)}}\\
= \dfrac{{2x + 6}}{{x\left( {2x + 6} \right)}}\\
= \dfrac{1}{x}
\end{array}$
