Giải thích các bước giải:
$DK:x \ne \left\{ {0; - 2; - 4; - 6; - 8} \right\}$
Ta có:
$\begin{array}{l}
\dfrac{1}{{{x^2} + 2x}} + \dfrac{1}{{{x^2} + 6x + 8}} + \dfrac{1}{{{x^2} + 10x + 24}} + \dfrac{1}{{{x^2} + 14x + 48}} = \dfrac{4}{{105}}\\
\Leftrightarrow \dfrac{1}{{x\left( {x + 2} \right)}} + \dfrac{1}{{\left( {x + 2} \right)\left( {x + 4} \right)}} + \dfrac{1}{{\left( {x + 4} \right)\left( {x + 6} \right)}} + \dfrac{1}{{\left( {x + 6} \right)\left( {x + 8} \right)}} = \dfrac{4}{{105}}\\
\Leftrightarrow \dfrac{2}{{x\left( {x + 2} \right)}} + \dfrac{2}{{\left( {x + 2} \right)\left( {x + 4} \right)}} + \dfrac{2}{{\left( {x + 4} \right)\left( {x + 6} \right)}} + \dfrac{2}{{\left( {x + 6} \right)\left( {x + 8} \right)}} = \dfrac{8}{{105}}\\
\Leftrightarrow \dfrac{1}{x} - \dfrac{1}{{x + 2}} + \dfrac{1}{{x + 2}} - \dfrac{1}{{x + 4}} + \dfrac{1}{{x + 4}} - \dfrac{1}{{x + 6}} + \dfrac{1}{{x + 6}} - \dfrac{1}{{x + 8}} = \dfrac{8}{{105}}\\
\Leftrightarrow \dfrac{1}{x} - \dfrac{1}{{x + 8}} = \dfrac{8}{{105}}\\
\Leftrightarrow \dfrac{8}{{x\left( {x + 8} \right)}} = \dfrac{8}{{105}}\\
\Rightarrow {x^2} + 8x - 105 = 0\\
\Leftrightarrow \left( {x - 7} \right)\left( {x + 15} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x = - 15
\end{array} \right.\left( c \right)
\end{array}$
Vậy phương trình có tập nghiệm là: $S = \left\{ { - 15;7} \right\}$
