Question

Va rog am nevoie de ajutor!!

Answer 1

a) 3x-2|x-5|≤3x-12=> |x-5|≥6
daca -6<x-5<6 atunci x ar fi apartinut intervalului (-1;11) 
  => x∈(-∞;-1]∪[11;+∞)

b) 3(2x+1)-x≤5x+7 <=> 6x+3-x≤5x+7 <=> 5x+3≤5x+7 <=> 3≤7 (A) => x∈R

c) (x+2)²-x(x+3)>x+4 <=> x²+4x+4-x²-3x>x+4 <=> x+4>x+4 <=> 4>4 (fals) => x∈∅

d) (x²+2x+3)(1-3x) <=0
   
      x²+2x+3=0
      Δ=4-12=-8<0 => x²+2x+3>0 indiferent de x
    
     atunci 1-3x≤0=> 3x≥1=> x≥$\frac{1}{3}$ 

     => x∈[$\frac{1}{3}$,+∞)