Đáp án: x=-4
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
{x^2} - 3x + 2 \ne 0\\
{x^2} - 4x + 3 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left( {x - 1} \right)\left( {x - 2} \right) \ne 0\\
\left( {x - 1} \right)\left( {x - 3} \right) \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne 2\\
x \ne 3
\end{array} \right.\\
\frac{{x + 4}}{{{x^2} - 3x + 2}} + \frac{{x + 1}}{{{x^2} - 4x + 3}} = \frac{{2x + 5}}{{{x^2} - 4x + 3}}\\
\Rightarrow \frac{{x + 4}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \frac{{2x + 5 - x - 1}}{{{x^2} - 4x + 3}}\\
\Rightarrow \frac{{x + 4}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} = \frac{{x + 4}}{{\left( {x - 1} \right)\left( {x - 3} \right)}}\\
\Rightarrow \left( {x + 4} \right)\left( {\frac{1}{{\left( {x - 1} \right)\left( {x - 2} \right)}} - \frac{1}{{\left( {x - 1} \right)\left( {x - 3} \right)}}} \right) = 0\\
\Rightarrow \left( {x + 4} \right)\frac{{x - 3 - \left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}} = 0\\
\Rightarrow \left[ \begin{array}{l}
x = - 4\left( {tm} \right)\\
x - 3 - x + 2 = 0\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = - 4
\end{array}$
