Question

Ajuuutooorr!!! Este urgent!! Exercițiul 4 punctul c !!!! Dau coroană!!

Answer 1

$x\in [-2,1] \Rightarrow x-1\in [-3,0]\Rightarrow |x-1|= 1-x\\x\in [-2,1] \Rightarrow x+2\in [0,3]\Rightarrow |x+2|=x+2\\\text{Prin urmare ecuatia va deveni :}\\\dfrac{|x|}{1-x+x+2}=0,(4)\\\dfrac{|x|}{3}=\dfrac{4}{9}\\|x|= 3\cdot \dfrac{4}{9}\\|x|=\dfrac{4}{3}\\x\in \left\{\pm\dfrac{4}{3}\right\}$

Answer 2

Poza conține rezolvarea.