Đáp án:

$\begin{array}{l}
a)\dfrac{{11}}{x} = \dfrac{9}{{x + 1}} + \dfrac{2}{{x - 4}}\left( {dkxd:\left\{ \begin{array}{l}
x \ne 0\\
x \ne  - 1\\
x \ne 4
\end{array} \right.} \right)\\
 \Rightarrow \dfrac{{11\left( {x + 1} \right)\left( {x - 4} \right)}}{{x\left( {x + 1} \right)\left( {x - 4} \right)}} = \dfrac{{9x\left( {x - 4} \right) + 2x\left( {x + 1} \right)}}{{x\left( {x + 1} \right)\left( {x - 4} \right)}}\\
 \Rightarrow 11\left( {{x^2} - 3x - 4} \right) = 9{x^2} - 36x + 2{x^2} + 2x\\
 \Rightarrow 11{x^2} - 33x - 44 = 11{x^2} - 34x\\
 \Rightarrow x = 44\left( {tmdk} \right)\\
b)Dkxd:x \ne 4\\
\dfrac{{14}}{{3x - 12}} - \dfrac{{2 + x}}{{x - 4}} = \dfrac{3}{{8 - 2x}} - \dfrac{5}{6}\\
 \Rightarrow \dfrac{{14}}{{3\left( {x - 4} \right)}} - \dfrac{{x + 2}}{{x - 4}} = \dfrac{{ - 3}}{{2\left( {x - 4} \right)}} - \dfrac{5}{6}\\
 \Rightarrow \dfrac{{14.4 - 12\left( {x + 2} \right)}}{{12\left( {x - 4} \right)}} = \dfrac{{ - 3.6 - 5.2.\left( {x - 4} \right)}}{{12\left( {x - 4} \right)}}\\
 \Rightarrow 64 - 12x - 24 =  - 18 - 10x + 40\\
 \Rightarrow 2x = 18\\
 \Rightarrow x = 9\left( {tmdk} \right)\\
c)Dkxd:x \ne \dfrac{1}{3};x \ne  - \dfrac{1}{3}\\
\dfrac{{12}}{{1 - 9{x^2}}} = \dfrac{{1 - 3x}}{{1 + 3x}} - \dfrac{{1 + 3x}}{{1 - 3x}}\\
 \Rightarrow \dfrac{{12}}{{\left( {1 + 3x} \right)\left( {1 - 3x} \right)}} = \dfrac{{{{\left( {1 - 3x} \right)}^2} - {{\left( {1 + 3x} \right)}^2}}}{{\left( {1 + 3x} \right)\left( {1 - 3x} \right)}}\\
 \Rightarrow 12 =  - 6x - 6x\\
 \Rightarrow x =  - 1\left( {tmdk} \right)\\
d)Dkxd:x \ne 0;x \ne 5;x \ne  - 5\\
\dfrac{{x + 5}}{{{x^2} - 5x}} - \dfrac{{x + 25}}{{2{x^2} - 50}} = \dfrac{{x - 5}}{{2{x^2} + 10x}}\\
 \Rightarrow \dfrac{{x + 5}}{{x\left( {x - 5} \right)}} - \dfrac{{x + 25}}{{2\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{x - 5}}{{2x\left( {x + 5} \right)}}\\
 \Rightarrow \dfrac{{2\left( {x + 5} \right)\left( {x + 5} \right) - x\left( {x + 25} \right) - {{\left( {x - 5} \right)}^2}}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}} = 0\\
 \Rightarrow 2{x^2} + 20x + 50 - {x^2} - 25x - {x^2} + 10x - 25 = 0\\
 \Rightarrow 5x =  - 25\\
 \Rightarrow x =  - 5\left( {ktm} \right)\\
 \Rightarrow x \in \emptyset 
\end{array}$