Đáp án:

\[\left[ \begin{array}{l}
x = 7\\
x =  - 15
\end{array} \right.\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\frac{1}{{{x^2} + 2x}} + \frac{1}{{{x^2} + 6x + 8}} + \frac{1}{{{x^2} + 10x + 24}} + \frac{1}{{{x^2} + 14x + 48}} = \frac{4}{{105}}\\
 \Leftrightarrow \frac{1}{{x\left( {x + 2} \right)}} + \frac{1}{{\left( {x + 2} \right)\left( {x + 4} \right)}} + \frac{1}{{\left( {x + 4} \right)\left( {x + 6} \right)}} + \frac{1}{{\left( {x + 6} \right)\left( {x + 8} \right)}} = \frac{4}{{105}}\\
 \Leftrightarrow \frac{2}{{x\left( {x + 2} \right)}} + \frac{2}{{\left( {x + 2} \right)\left( {x + 4} \right)}} + \frac{2}{{\left( {x + 4} \right)\left( {x + 6} \right)}} + \frac{2}{{\left( {x + 6} \right)\left( {x + 8} \right)}} = \frac{8}{{105}}\\
 \Leftrightarrow \frac{{\left( {x + 2} \right) - x}}{{x\left( {x + 2} \right)}} + \frac{{\left( {x + 4} \right) - \left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x + 4} \right)}} + \frac{{\left( {x + 6} \right) - \left( {x + 4} \right)}}{{\left( {x + 4} \right)\left( {x + 6} \right)}} + \frac{{\left( {x + 8} \right) - \left( {x + 6} \right)}}{{\left( {x + 6} \right)\left( {x + 8} \right)}} = \frac{8}{{105}}\\
 \Leftrightarrow \frac{1}{x} - \frac{1}{{x + 2}} + \frac{1}{{x + 2}} - \frac{1}{{x + 4}} + \frac{1}{{x + 4}} - \frac{1}{{x + 6}} + \frac{1}{{x + 6}} - \frac{1}{{x + 8}} = \frac{8}{{105}}\\
 \Leftrightarrow \frac{1}{x} - \frac{1}{{x + 8}} = \frac{8}{{105}}\\
 \Leftrightarrow \frac{{x + 8 - x}}{{x\left( {x + 8} \right)}} = \frac{8}{{105}}\\
 \Leftrightarrow \frac{8}{{x\left( {x + 8} \right)}} = \frac{8}{{105}}\\
 \Leftrightarrow x\left( {x + 8} \right) = 105\\
 \Leftrightarrow {x^2} + 8x - 105 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x =  - 15
\end{array} \right.
\end{array}\)

Đáp án:

\(\left[ \begin{array}{l}x=7\\x=-15\end{array} \right.\) 

Giải thích các bước giải:

 PT đã cho ⇔ $\frac{1}{x(x+2)}$ +$\frac{1}{(x+2)(x+4)}$ +$\frac{1}{(x+4)(x+6)}$ +$\frac{1}{(x+6)(x+8)}$ =$\frac{4}{105}$

⇔ $\frac{2}{x(x+2)}$ +$\frac{2}{(x+2)(x+4)}$ +$\frac{2}{(x+4)(x+6)}$ +$\frac{2}{(x+6)(x+8)}$ =$\frac{8}{105}$

⇔ $\frac{1}{x}$ -$\frac{1}{x+2}$ +$\frac{1}{x+2}$-$\frac{1}{x+4}$ +...-$\frac{1}{x+8}$ =$\frac{8}{105}$ 

⇔ $\frac{1}{x}$ -$\frac{1}{x+8}$ =$\frac{8}{105}$ 

⇔ $\frac{8}{x(x+8)}$ =$\frac{8}{105}$ 

⇔ $x^{2}$ +$8x^{}$-$105^{}$ =$0^{}$ 

⇔ \(\left[ \begin{array}{l}x=7\\x=-15\end{array} \right.\) .