Đáp án:
$\begin{array}{l}
a)\dfrac{2}{x} = \dfrac{{5 + 4x}}{2}\\
\Rightarrow x.\left( {5 + 4x} \right) = 2.2\\
\Rightarrow 4{x^2} + 5x - 4 = 0\\
\Rightarrow {\left( {2x} \right)^2} + 2.2x.\dfrac{5}{4} + \dfrac{{25}}{{16}} = \dfrac{{41}}{{16}}\\
\Rightarrow {\left( {2x + \dfrac{5}{4}} \right)^2} = \dfrac{{41}}{{16}}\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 5 + \sqrt {41} }}{8}\\
x = \dfrac{{ - 5 - \sqrt {41} }}{8}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 5 \pm \sqrt {41} }}{8}\\
b)\dfrac{{x + 2}}{x} = \dfrac{{ - 2x - 3}}{{2\left( {2 - x} \right)}}\\
\Rightarrow \left( {x + 2} \right).2.\left( {2 - x} \right) = x.\left( { - 2x - 3} \right)\\
\Rightarrow 2.\left( {4 - {x^2}} \right) = - 2{x^2} - 3x\\
\Rightarrow - 2{x^2} + 8 = - 2{x^2} - 3x\\
\Rightarrow - 3x = 8\\
\Rightarrow x = \dfrac{{ - 8}}{3}\\
Vậy\,x = \dfrac{{ - 8}}{3}\\
c)Dk:x \ne - 2\\
1 + \dfrac{1}{{x + 2}} = \dfrac{{12}}{{\left( {x + 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
\Rightarrow \dfrac{{\left( {x + 2} \right)\left( {{x^2} + 2x + 4} \right) + {x^2} + 2x + 4}}{{\left( {x + 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{12}}{{\left( {x + 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
\Rightarrow {x^3} + 4{x^2} + 8x + 8 + {x^2} + 2x + 4 = 12\\
\Rightarrow {x^3} + 5{x^2} + 12x = 0\\
\Rightarrow x\left( {{x^2} + 5x + 12} \right) = 0\\
\Rightarrow x = 0\left( {tm} \right)\\
Vậy\,x = 0
\end{array}$
