Question

Exercițiul 3
Dau coroana!

Answer 1

3x-2 /5x+2 >=0 ;conditia de existenta este 5x+2≠0 <=> x≠-2/5 iar cazul de egalitate are loc pentru 3x-2=0 <=> x=2/3 .

Daca 3x-2 /5x+2 >0 => 3x-2 >0 <=> x >2/3 dar x∈R <=> x∈(2/3,∞) si 5x+2 >0 <=> x >-2/5 dar x∈R <=> x∈(-2/5,∞) => x∈(-2/5,∞) ∩ (2/3,∞) =(2/3,∞) .

sau 3x-2 <0 <=> x <2/3 dar x∈R <=> x∈(-∞,2/3) si 5x+2 <0 <=> x < -2/5 dar x∈R <=> x∈(-∞,-2/5) => x∈(-∞,2/3) ∩ (-∞,-2/5) =(-∞,-2/5) .

Asadar S={ 2/3 } ∪ (2/3,∞) ∪ (-∞,-2/5) .