Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne - 3\\
\dfrac{{2x - 3}}{{x + 3}} = 1\\
\Rightarrow 2x - 3 = x + 3\\
\Rightarrow 2x - x = 3 + 3\\
\Rightarrow x = 6\left( {tmdk} \right)\\
Vậy\,x = 6\\
b)Dkxd:x \ne 2\\
\dfrac{{\left( {{x^2} + 2x} \right) - \left( {2x + 4} \right)}}{{x - 2}} = 0\\
\Rightarrow {x^2} + 2x - 2x - 4 = 0\\
\Rightarrow {x^2} = 4\\
\Rightarrow \left[ \begin{array}{l}
x = 2\left( {ktm} \right)\\
x = - 2\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = - 2\\
c)Dkxd:x \ne 2\\
\dfrac{{4x + 2}}{{3x - 6}} - \dfrac{x}{{2 - x}} = 1 + \dfrac{{3x}}{{2x - 4}}\\
\Rightarrow \dfrac{{4x + 2}}{{3\left( {x - 2} \right)}} + \dfrac{x}{{x - 2}} = 1 + \dfrac{3}{{2\left( {x - 2} \right)}}\\
\Rightarrow \dfrac{{2.\left( {4x + 2} \right) + 6.x}}{{6\left( {x - 2} \right)}} = \dfrac{{6\left( {x - 2} \right) + 3.3}}{{6\left( {x - 2} \right)}}\\
\Rightarrow 8x + 4 + 6x = 6x - 12 + 9\\
\Rightarrow 8x = - 7\\
\Rightarrow x = - \dfrac{7}{8}\left( {tmdk} \right)\\
Vậy\,x = - \dfrac{7}{8}\\
d)Dkxd:x \ne 3;x \ne - 3\\
\dfrac{{x - 3}}{{x + 3}} - \dfrac{{x + 3}}{{x - 3}} = \dfrac{3}{{{x^2} - 9}}\\
\Rightarrow \dfrac{{{{\left( {x - 3} \right)}^2} - {{\left( {x + 3} \right)}^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{3}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
\Rightarrow - 6x = 3\\
\Rightarrow x = - \dfrac{1}{2}\left( {tmdk} \right)\\
Vậy\,x = - \dfrac{1}{2}
\end{array}$
