Question

Am |x-|x-1|+1|<=2. Cum fac? E urgent, vă rog mult.​

Answer 1

Metoda I:

$|x-|x-1|+1|\leq 2$

$\Rightarrow -2\leq x-|x-1|+1\leq 2\big|-x-1$

① $x-1\geq 0\Rightarrow x\geq 1$:

$\Rightarrow -2\leq x-(x-1)+1\leq 2$

$\Rightarrow -2\leq 2\leq 2\,\,\,\text{(A)}$

$\Rightarrow x\geq 1 \Rightarrow x\in [1,+\infty)$

② $x-1<0\Rightarrow x<1$:

$\Rightarrow -2\leq x+(x-1)+1\leq 2$

$\Rightarrow -2\leq 2x\leq 2$

$\Rightarrow -1\leq x\leq 1$

$\Rightarrow (-1\leq x\leq 1)\, \wedge\, (x <1)\Rightarrow x\in [-1,1)$

Din ① ∨ ②:

$\Rightarrow x\in [1,+\infty)\cup [-1,1) \Rightarrow \boxed{x\in [-1,+\infty)}$

Metoda II:

$|x-|x-1|+1|\leq 2$

$\Rightarrow -2\leq x-|x-1|+1\leq 2\big|-x-1$

$\Rightarrow -x-3\leq -|x-1|\leq -x+1\big|\cdot (-1)$

$\Rightarrow x+3\geq |x-1|\geq x-1\\ \\ \Rightarrow \left\{\begin{aligned}|x-1|\leq x+3\\ |x-1|\geq x-1\end{aligned}\right.\Rightarrow \left\{\begin{aligned}&(-x-3\leq x-1)\,\wedge\,(x-1\leq x+3)\\ &\big[-(x-1)\geq x-1\big]\,\vee\,(x-1\geq x-1)\end{aligned}\right.$

$\Rightarrow \left\{\begin{aligned}&(-2\leq 2x)\,\wedge\,(-1\leq 3)\\ &(2\geq 2x)\,\vee\,(-1\geq -1)\end{aligned}\right.\Rightarrow \left\{\begin{aligned}&(x\geq -1)\, \wedge\,(x\in \mathbb{R})\\ &(x\leq 1)\,\vee\,(x\in \mathbb{R})\end{aligned}\right.$

$\Rightarrow \left\{\begin{aligned}&x\geq -1\\ &x\in \mathbb{R}\end{aligned}\,\right|\Rightarrow x\geq -1 \Rightarrow \boxed{x\in [-1,+\infty)}$