Question

Inecuatie cu modul.
|x| > |x+1|

Answer 1

|x| >|x+1|
|x| - |x+1| >  0

x-(x+1)>0 ,x ≥ 0 ,x+1 ≥0 
-x-(x+1)>0 ,x <0, x+1≥0
x-(-(x+1)>0, x≥0 x+1<0
-x-(-(x+1)) >0 ,x<0 , x+1<0
x ∈ Ø ,x≥0, x≥-1
x< - 1/2, x<0 , x≥ -1
x > - 1/2, x∈Ø 
x ∈ R , x∈(-∞,-1)

x ∈ Ø 
x ∈ (-1,-1/2)
x ∈ (-∞,-1)

deci x ∈ (-∞,-1/2)

Answer 2

$|x|\ \textgreater \ |x+1| \Rightarrow |x|-|x+1|\ \textgreater \ 0 \\ \\ |x|-|x+1| =\left\{ \begin{array}{c} -x-\big(-(x+1)\big),\quad x\ \textless \ -1\\ 1, \quad x=-1 \\ -x-(x+1),\quad x\in(-1,0) \\ -1,\quad x=0\\ x-(x+1), \quad x\ \textgreater \ 0 \end{array} \right|\\ \\ \boxed{1}\quad Daca x\ \textless \ -1: \\ \\ -x-\big(-(x+1)\big)\ \textgreater \ 0 \Rightarrow -x+x+1\ \textgreater \ 0 \Rightarrow 1\ \textgreater \ 0,\quad \forall x\in \mathbb_{R} \\ \\ Dar x\ \textless \ -1 \Rightarrow\boxed{ x\in (-\infty,-1)}\\ \\ \boxed{2} \quad Daca x=-1:$
$1 \ \textgreater \ 0,\quad \forall x\in \mathbb_{R}, Dar, x=-1 \Rightarrow \boxed{x \in \Big\{-1\Big\}} \\ \\ \boxed{3}\quad Daca x\in(-1,0): \\ \\ -x-(x+1)\ \textgreater \ 0 \Rightarrow -x-x-1\ \textgreater \ 0 \Rightarrow -2x-1\ \textgreater \ 0 \Rightarrow \\ \\ \Rightarrow -2x\ \textgreater \ 1 \Big|\cdot (-1) \Rightarrow 2x<-1 \Rightarrow x<-\dfrac{1}{2}, Dar x \in (-1,0) \Rightarrow \\ \\ \Rightarrow \boxed{x\in \Big(-1,-\dfrac{1}{2}\Big)}$

$\boxed{4}\quad Daca x=0: \\ \\ -1\ \textgreater \ 0 (F) \Rightarrow \boxed{x\in \emptyset}$

$\boxed{5} \quad Daca x\ \textgreater \ 0:\\ \\ x-(x+1) \ \textgreater \ 0 \Rightarrow x-x-1\ \textgreater \ 0 \Rightarrow -1\ \textgreater \ 0 \Rightarrow \boxed{x\in \emptyset} \\ \\ \\ Din \boxed{1} \cup \boxed{2} \cup \boxed{3} \cup \boxed{4} \cup \boxed{5} \Rightarrow \boxed{S = \Big(-\infty, -\dfrac{1}{2}\Big)}$