Răspuns:
Explicație pas cu pas:
3(2x-1) - 15 ≤ x + 32 <=>
6x-3 -15 ≤ x +32 =>
6x-x ≤ 32+18 =>
5x ≤ 40 => x ≤ 40:5 => x ≤ 8 => x ∈ -∞ ; 8]
²⁾(x-1)/3 + ³⁾(2x-5)/2 ≤ (3x-2)/6 =>
2(x-1) + 3(2x-5) ≤ 3x-2 <=>
2x-2 +6x-15 ≤ 3x-2 =>
8x-3x ≤ 17-2 => 5x ≤ 15 I:5 => x ≤ 3 => x ∈ (-∞ ; 3]
5x - 2(x+7) ≤ 5(2x+5) + 3 <=>
5x - 2x-14 ≤ 10x + 25 + 3 <=>
3x -14 ≤ 10x +28 =>
-14-28 ≤ 10x-3x =>
-42 ≤ 7x I:7 =>
-6 ≤ x => x ≥ -6 => x ∈ [-6 ; +∞)
