ĐKXĐ: \(\left\{{}\begin{matrix}x^2-3x+2\ne0\\x^2-4x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne3\end{matrix}\right.\)
\(\dfrac{x+4}{x^2-3x+2}-\dfrac{x+1}{x^2-4x+3}=\dfrac{2x+5}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{x+4}{x^2-3x+2}-\dfrac{x+1}{x^2-4x+3}-\dfrac{2x+5}{x^2-4x+3}=0\)
\(\Leftrightarrow\dfrac{x+4}{x^2-2x-x+2}-\dfrac{3x+6}{x^2-3x-x-3}=0\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}-\dfrac{3x+6}{\left(x-3\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}-\dfrac{3\left(x+2\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+x-12-3x^2+12}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{-2x^2+x}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{-x\left(2x-1\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0;\dfrac{1}{2}\right\}\)