Đáp án:
$\begin{array}{l}
d)Dkxd:x \ne \pm 1;x \ne \pm 2\\
\frac{{x + 4}}{{x - 1}} + \frac{{x - 4}}{{x + 1}} = \frac{{x + 8}}{{x - 2}} + \frac{{x - 8}}{{x + 2}} + 6\\
\Rightarrow \frac{{x - 1 + 5}}{{x - 1}} + \frac{{x + 1 - 5}}{{x + 1}} = \frac{{x - 2 + 10}}{{x - 2}} + \frac{{x + 2 - 10}}{{x + 2}} + 6\\
\Rightarrow 1 + \frac{5}{{x - 1}} + 1 - \frac{5}{{x + 1}} = 1 + \frac{{10}}{{x - 2}} + 1 - \frac{{10}}{{x + 2}} + 6\\
\Rightarrow 5\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) - 10.\left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}}} \right) = 6\\
\Rightarrow 5.\frac{{x + 1 - x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - 10.\frac{{x + 2 - x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 6\\
\Rightarrow \frac{{10}}{{{x^2} - 1}} - \frac{{40}}{{{x^2} - 4}} = 6\\
\Rightarrow \frac{{5\left( {{x^2} - 4} \right) - 20\left( {{x^2} - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^2} - 4} \right)}} = 3\\
\Rightarrow - 15{x^2} = 3\left( {{x^4} - 5{x^2} + 4} \right)\\
\Rightarrow {x^4} - 5{x^2} + 4 = - 5{x^2}\\
\Rightarrow {x^4} + 4 = 0\\
\Rightarrow {x^4} = - 4
\end{array}$
Vậy pt vô nghiệm.
