Question

Sa se gaseasca multimea tuturor valorilor lui x ∈ R astfel incat:
$\sqrt{(x^2-3x+2} \ \textgreater \ x + 1$

Answer 1

$\sqrt{x^2-3x+2}>x+1\\\\ \text{Conditii de existenta:} \\ x^2-3x+2 \geq 0 \Rightarrow (x-1)(x-2) \geq 0 \Rightarrow D= (-\infty,1]\cup[2,+\infty) \\ \\ \sqrt{x^2-3x+2}>x+1\Big|^2,\quad x+1 > 0 \\ x^2-3x+2 > x^2+2x+1 \\ -5x > -1 \\ 5x<1 \\ \\x < \dfrac{1}{5} \\ \\ \Rightarrow x\in \Big(-\infty, \dfrac{1}{5}\Big)\,\cap D \, \Leftrightarrow\, \Big(-\infty, \dfrac{1}{5}\Big)\,\cap \Big( (-\infty,1]\cup[2,+\infty)\Big) \Rightarrow \\ \\ \\\Rightarrow \boxed{x\in \Big(-\infty,\dfrac{1}{5}\Big)}$