Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne - 3\\
\dfrac{{2x - 3}}{{x + 3}} = 1\\
\Rightarrow 2x - 3 = x + 3\\
\Rightarrow 2x - x = 3 + 3\\
\Rightarrow x = 6\left( {tmdk} \right)\\
Vậy\,x = 6\\
b)Dkxd:x \ne 0\\
\dfrac{{{x^2} - 4}}{x} = x + 2\\
\Rightarrow {x^2} - 4 = x.\left( {x + 2} \right)\\
\Rightarrow {x^2} - 4 = {x^2} + 2x\\
\Rightarrow 2x = - 4\\
\Rightarrow x = - 2\left( {tmdk} \right)\\
Vậy\,x = - 2\\
c)Dkxd:x \ne 1\\
\dfrac{{5x - 1}}{{x - 1}} + 3 = 1 - \dfrac{x}{{x - 1}}\\
\Rightarrow \dfrac{{5x - 1 + 3\left( {x - 1} \right)}}{{x - 1}} = \dfrac{{x - 1 - x}}{{x - 1}}\\
\Rightarrow 5x - 1 + 3x - 3 = - 1\\
\Rightarrow 8x = 3\\
\Rightarrow x = \dfrac{3}{8}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{3}{8}
\end{array}$
