Đáp án:
$\begin{array}{l}
Dkxd:x \ne 0;x \ne 2;x \ne - 2\\
\frac{2}{{{x^2} - 4}} - \frac{{x - 1}}{{x\left( {x - 2} \right)}} + \frac{{x - 4}}{{x\left( {x + 2} \right)}} = 0\\
\Rightarrow \frac{2}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \frac{{x - 1}}{{x\left( {x - 2} \right)}} + \frac{{x - 4}}{{x\left( {x + 2} \right)}} = 0\\
\Rightarrow \frac{{2x - \left( {x - 1} \right).\left( {x + 2} \right) + \left( {x - 4} \right)\left( {x - 2} \right)}}{{x\left( {x + 2} \right)\left( {x - 2} \right)}} = 0\\
\Rightarrow 2x - {x^2} - x + 2 + {x^2} - 6x + 8 = 0\\
\Rightarrow - 5x + 10 = 0\\
\Rightarrow x = 2\left( {ktm} \right)\\
\Rightarrow pt\,vô\,nghiệm\\
Dkxd:x \ne - 1;x \ne 3\\
\frac{1}{3} - \frac{{x - 1}}{{x + 1}} = \frac{x}{{x - 3}} - \frac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} - 2x - 3}}\\
\Rightarrow \frac{1}{3} - \frac{{x - 1}}{{x + 1}} = \frac{x}{{x - 3}} - \frac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\\
\Rightarrow \frac{{\left( {x + 1} \right)\left( {x - 3} \right) - 3.\left( {x - 1} \right)\left( {x - 3} \right)}}{{\left( {x + 1} \right)\left( {x - 3} \right)}} = \frac{{3x\left( {x + 1} \right) - 3{{\left( {x - 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\\
\Rightarrow {x^2} - 2x - 3 - 3{x^2} + 12x - 9\\
= 3{x^2} + 3x - 3{x^2} + 6x - 3\\
\Rightarrow - 2{x^2} + x - 9 = 0\\
\Rightarrow 2{x^2} - x + 9 = 0\\
\Rightarrow 2\left( {{x^2} - \frac{1}{2}x + \frac{1}{{16}}} \right) + \frac{{71}}{8} = 0\\
\Rightarrow 2{\left( {x - \frac{1}{4}} \right)^2} + \frac{{71}}{8} = 0\left( {vn} \right)
\end{array}$
Vậy phương trình vô nghiệm
