Question

|x-5|+|x-2|-1=0
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ajutor

Answer 1

$|x-5|+|x-2|-1 = \left\{ \begin{array}{c} -(x-5)-(x-2)-1,\quad x\in(-\infty,2] \quad \boxed{I}\\ -(x-5)+x-2-1, \quad x\in(2,5]\quad \boxed{II}\\ x-5+x-2-1,\quad x\in(5,\infty)\quad \boxed{III} \end{array} \right \\ \\ \\ \boxed{1} Pentru x\in(-\infty,2]: \\ -(x-5)-(x-2)-1 = 0 \Rightarrow -x+5-x+2-1 = 0 \Rightarrow \\ -2x +6 = 0 \Rightarrow -2x = -6 \Rightarrow x = 3 \notin (-\infty, 2]$

$\boxed{2} Pentru x\in (2,5]: \\ -(x-5)+x-2-1 = 0 \Rightarrow -x+5+x-3 = 0 \Rightarrow 2 = 0 \Rightarrow x\in \O$

$\boxed{3} Pentru x\in(5,\infty):\\ x-5+x-2-1 = 0 \Rightarrow 2x - 8 = 0 \Rightarrow 2x = 8 \Rightarrow x = 4 \notin [5,\infty)\quad \\ \\ \\ Din \boxed{1} \cup \boxed{2} \cup \boxed{3} \Rightarrow \boxed{\boxed{S = \Big\{\O\Big\}}}$