Đáp án:
$S = \left\{ {\dfrac{{ - 19 - \sqrt {505} }}{{12}};\dfrac{6}{5}} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
\dfrac{{2x + 3}}{{x + 3}} + \dfrac{{3x - 2}}{{\left| x \right|}} = 5\left( {DK:x \ne \left\{ { - 3;0} \right\}} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
\dfrac{{2x + 3}}{{x + 3}} + \dfrac{{3x - 2}}{x} = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
\dfrac{{2x + 3}}{{x + 3}} + \dfrac{{3x - 2}}{{ - x}} = 5
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
\left( {2x + 3} \right)x + \left( {3x - 2} \right)\left( {x + 3} \right) = 5x\left( {x + 3} \right)
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
\left( {2x + 3} \right)\left( { - x} \right) + \left( {3x - 2} \right)\left( {x + 3} \right) = 5\left( { - x} \right)\left( {x + 3} \right)
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
5{x^2} + 10x - 6 = 5{x^2} + 15x
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
{x^2} + 4x - 6 = - 5{x^2} - 15x
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
5x = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
6{x^2} + 19x - 6 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x = \dfrac{6}{5}\left( {tm} \right)
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
\left[ \begin{array}{l}
x = \dfrac{{ - 19 + \sqrt {505} }}{{12}}(ktm)\\
x = \dfrac{{ - 19 - \sqrt {505} }}{{12}}\left( {tm} \right)
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{6}{5}\\
x = \dfrac{{ - 19 - \sqrt {505} }}{{12}}
\end{array} \right.
\end{array}$
Vậy tập nghiệm của phương trình là $S = \left\{ {\dfrac{{ - 19 - \sqrt {505} }}{{12}};\dfrac{6}{5}} \right\}$
