ĐKXĐ: \(x\ne-3;-2;-1;0\)
\(\frac{5}{x}+\frac{2}{x+3}=\frac{4}{x+1}+\frac{3}{x+2}\)
\(\Leftrightarrow\left(\frac{5}{x}+1\right)+\left(\frac{2}{x+3}+1\right)=\left(\frac{4}{x+1}+1\right)+\left(\frac{3}{x+2}+1\right)\)
\(\Leftrightarrow\frac{x+5}{x}+\frac{x+5}{x+3}-\frac{x+5}{x+1}-\frac{x+5}{x+2}=0\)
\(\Leftrightarrow\left(x+5\right)\left(\frac{1}{x}+\frac{1}{x+3}-\frac{1}{x+1}-\frac{1}{x+2}\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left[\frac{x+3+x}{x\left(x+3\right)}-\frac{x+2+x+1}{\left(x+1\right)\left(x+2\right)}\right]=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+3\right)\left(\frac{1}{x\left(x+3\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)=0\)
Vì \(x\left(x+3\right)-\left(x+1\right)\left(x+2\right)=x^2+3x-x^2-3x-2=-2< 0\)
\(\Rightarrow\frac{1}{x\left(x+3\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\ne0\)
Vậy: \(\left[{}\begin{matrix}x+5=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\frac{3}{2}\end{matrix}\right.\) (thoả)
Vậy \(S=\left\{-5;-\frac{3}{2}\right\}\)
