[tex]\sqrt{x^2-3x+2}>x+1\\ \\ D.V.A.:\quad x^2-3x+2 \geq 0 \Rightarrow (x-2)(x-1) \geq 0 \Rightarrow\\\\\Rightarrow x\in (-\infty, 1]\cup[2,+\infty)\\ \\\\ \boxed{1}\quad x+1 \geq 0 \Rightarrow x\geq -1\\ \\ \sqrt{x^2-3x+2}>x+1 \Big|^2 \\ x^2-3x+2>(x+1)^2 \\ x^2-3x+2 > x^2+2x+1\\ -5x > -1 \\ x < \dfrac{1}{5} \Rightarrow x\in \Big[-1,\dfrac{1}{5}\Big)\\ \\ \\\boxed{2}\quad x+1 < 0 \Rightarrow x< -1\\ \\ \sqrt{x^2-3x+2}>x+1 \quad (A)\\\\ \Rightarrow x<-1 \Rightarrow x\in (-\infty, -1)[/tex]
[tex]\text{Din }\boxed{1}\,\cup\,\boxed{2} \Rightarrow x\in \Bigg(\Big[-1,\dfrac{1}{5}\Big)\cup(-\infty, -1)\Bigg)\cap D.V.A.\Rightarrow \\ \\ \Rightarrow \boxed{x\in \Big(-\infty, \dfrac{1}{5}\Big)}[/tex]
