Giải thích các bước giải:
ĐKXĐ:$x \ne \left\{ { - 4; - 3; - 2; - 1} \right\}$
Ta có:
$\begin{array}{l}
\dfrac{1}{{x + 1}} + \dfrac{1}{{x + 2}} - \dfrac{1}{{x + 3}} - \dfrac{1}{{x + 4}} = \dfrac{{11}}{{12}}\\
\Leftrightarrow \left( {\dfrac{1}{{x + 1}} - \dfrac{1}{{x + 4}}} \right) + \left( {\dfrac{1}{{x + 2}} - \dfrac{1}{{x + 3}}} \right) = \dfrac{{11}}{{12}}\\
\Leftrightarrow \dfrac{3}{{\left( {x + 1} \right)\left( {x + 4} \right)}} + \dfrac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}} = \dfrac{{11}}{{12}}\\
\Leftrightarrow \dfrac{3}{{{x^2} + 5x + 4}} + \dfrac{1}{{{x^2} + 5x + 6}} = \dfrac{{11}}{{12}}\left( 1 \right)
\end{array}$
Đặt $t = {x^2} + 5x + 5$
Khi đó:
$\begin{array}{l}
\left( 1 \right)tt:\dfrac{3}{{t - 1}} + \dfrac{1}{{t + 1}} = \dfrac{{11}}{{12}}\\
\Leftrightarrow \dfrac{{4t + 2}}{{{t^2} - 1}} = \dfrac{{11}}{{12}}\\
\Leftrightarrow 11\left( {{t^2} - 1} \right) - 12\left( {4t + 2} \right) = 0\\
\Leftrightarrow 11{t^2} - 48t - 35 = 0\\
\Leftrightarrow \left( {t - 5} \right)\left( {11t + 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = 5\\
t = \dfrac{{ - 7}}{{11}}
\end{array} \right.
\end{array}$
$ + )TH1:t = 5$
Khi đó:
$\begin{array}{l}
{x^2} + 5x + 5 = 5\\
\Leftrightarrow {x^2} + 5x = 0\\
\Leftrightarrow x\left( {x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 5
\end{array} \right.\left( {tm} \right)
\end{array}$
$ + )TH2:t = \dfrac{{ - 7}}{{11}}$
Khi đó:
$\begin{array}{l}
{x^2} + 5x + 5 = \dfrac{{ - 7}}{{11}}\\
\Leftrightarrow {x^2} + 5x + \dfrac{{62}}{{11}} = 0\\
\Leftrightarrow {\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{{27}}{{44}} = 0\\
\Leftrightarrow {\left( {x + \dfrac{5}{2}} \right)^2} = \dfrac{{27}}{{44}}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 5}}{2} + \sqrt {\dfrac{{27}}{{44}}} \\
x = \dfrac{{ - 5}}{2} - \sqrt {\dfrac{{27}}{{44}}}
\end{array} \right.\left( {tm} \right)
\end{array}$
Vậy phương trình có tập nghiệm $S = \left\{ {0;5;\dfrac{{ - 5}}{2} + \sqrt {\dfrac{{27}}{{44}}} ;\dfrac{{ - 5}}{2} - \sqrt {\dfrac{{27}}{{44}}} } \right\}$
