Question

Rezolva inecuatiile:

1) x²+3x+2≥0
2) 2x≤ x²
3) 4x²≤1
4) (x-3)(x+7)≥0
5) x²-x+3≥0

Answer 1

folosim metoda intervalelor
1) x²+3x+2≥0 
(x+2)(x+1)≥0
x∈(-∞;-2]∪[-1;∞)

2) 2x≤ x²
x²-2x≥0
x(x-2)≥0
x∈(-∞;0]∪[2;∞)

3) 4x²≤1
4x²-1≤0
(2x-1)(2x+1)≤0
x∈[-1/2;+1/2]

4) (x-3)(x+7)≥0
x∈(-∞;-7]∪[3;∞)

5) x²-x+3≥0
x∈(-∞;∞)