4x (x-2) ≤5;

(2x+1) (x+1) ≥3;

3x (1-x) ≤-6;

(1-2x) (1-3x) ≤2

Question

Алгебра

Гемина

27 мая, 14:24

4x (x-2) ≤5;

(2x+1) (x+1) ≥3;

3x (1-x) ≤-6;

(1-2x) (1-3x) ≤2

+2

Ответы (1)

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Венюся · Answer 1

4x^2 - 8x < = 5

4x^2 - 8x - 5 < = 0

4x^2 - 8x - 5 = 0

D = 64+80=144 = 12^2

x1 = (8+12) / 8 = 2,5

x2 = (8-12) / 8 = - 1/2

2x^2 + x + x + 1 > = 3

2x^2 + 2x - 2 > = 0

x^2 + x - 1 > = 0

D = 1 + 4 = 5

x1 = (-1 + sqrt5) / 2

x2 = (-1 - sqrt5) / 2

4x (x-2) ≤5;(2x+1) (x+1) ≥3;3x (1-x) ≤-6;(1-2x) (1-3x) ≤2

4x (x-2) ≤5;

(2x+1) (x+1) ≥3;

3x (1-x) ≤-6;

(1-2x) (1-3x) ≤2