Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 2;x \ne - 2\\
\frac{{x + 2}}{{x - 2}} - \frac{{x - 2}}{{x + 2}} = \frac{4}{{{x^2} - 4}}\\
\Leftrightarrow \frac{{x + 2}}{{x - 2}} - \frac{{x - 2}}{{x + 2}} = \frac{4}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
\Leftrightarrow \frac{{\left( {x + 2} \right)\left( {x + 2} \right) - \left( {x - 2} \right)\left( {x - 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}} = \frac{4}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
\Rightarrow {x^2} + 4x + 4 - {x^2} + 4x - 4 = 4\\
\Rightarrow 8x = 4\\
\Rightarrow x = \frac{1}{2}\left( {tm} \right)\\
b)Dkxd:x \ne 1;x \ne - 3\\
\frac{{x + 1}}{{x - 1}} - \frac{{x + 2}}{{x + 3}} + \frac{4}{{{x^2} + 2x - 3}} = 0\\
\Leftrightarrow \frac{{x + 1}}{{x - 1}} - \frac{{x + 2}}{{x + 3}} + \frac{4}{{\left( {x - 1} \right)\left( {x + 3} \right)}} = 0\\
\Leftrightarrow \frac{{\left( {x + 1} \right)\left( {x + 3} \right) - \left( {x + 2} \right)\left( {x - 1} \right) + 4}}{{\left( {x - 1} \right)\left( {x + 3} \right)}} = 0\\
\Rightarrow {x^2} + 4x + 3 - {x^2} - x + 2 + 4 = 0\\
\Rightarrow 3x + 9 = 0\\
\Rightarrow x = - 3\left( {ktm} \right)
\end{array}$
Vậy phương trình vô nghiệm.
