[tex]\dfrac{x-3}{x+1}\ \geq\ \dfrac{2x-5}{x+2}\Leftrightarrow \dfrac{x-3}{x+1}\ -\ \dfrac{2x-5}{x+2}\ \geq\ 0\Leftrightarrow
\\\;\\
\Leftrightarrow \dfrac{(x-3)(x+2) - (x+1)(2x-5)}{(x+1)(x+2)} \geq 0[/tex]
Dupa efectuarea calculelor de la numarator, obtinem:
[tex]\dfrac{-(x-1)^2}{(x+1)(x+2)}\geq0|_{\cdot\ (-1)} \Leftrightarrow \dfrac{(x-1)^2}{(x+1)(x+2)}\leq 0\Leftrightarrow \begin{cases} (x+1)(x+2)\ \ \textless \ 0\\\;\\
x-1 = 0 \end{cases}
\\\;\\
S\ =\ (-1,\ -2) \cup\ \{1\}[/tex]
