§ 5: 5.5

5.5 Розв’яжіть нерівність:

x2- 4x2- 9 > 0, х-2(x+2)x-3(x+3) > 0,x ∈ (–∞; –3) ∪ (–2; 2) ∪ (3; +∞)6) x2- 3xx2- 8x+7 ≤ 0, x(x-3)x-7(x-1) ≤ 0,x ∈ [0; 1) ∪ [3; 7);

7) 2x2- 5x+2x2- 8x-4 ≥ 0, 1) 2x² – 5x + 2 = 0, D = 25 – 16 = 9,

2x-2(x-0,5)x-4(x+1) ≥ 0,x₁ = 5+34 = 2, x₂ = 5-34 = 12 = 0,5; 2) x² – 3x – 4 = 0,

x₁ + x₂ = 3, x = 4,

x₁ • x₂ = –4; x = –1;

x ∈ (–∞; –1) ∪ [0,5; 2] ∪ (4; +∞);