[tex]\dfrac{x^2-4x+3}{x^2+5x+6}\geq 0\\\\ \\ \Rightarrow \dfrac{(x-1)(x-3)}{(x+2)(x+3)}\geq 0\\ \\\\ \text{Din tabelul de variatie din imaginea atasata:}\\\\ \\ \Rightarrow \boxed{ x\in \big(-\infty, -3\big)\cup \big(-2, 1\big]\cup\big[3,+\infty\big)}[/tex]
