rozwiaż:
x(1-3x)<0
4x²≤ 5x
x<6x²
(-x+5)(x+2)>0
(1+x)(3-2x) ≤ 0
x²<1
3x²-1>2½ x
x-7≥5x
½ x²-x ≥ 1
4x²+49≤28x


Odpowiedź :

x(1-3x) < 0

1-3x < 0

-3x < -1 |:(-3)

x > ⅓
x∈(0,⅓)

4x² ≤ 5x
4x²-5x ≤ 0
x(4x-5) ≤ 0
4x-5=0  4x=5 x=1¼
x∈ <0, 1¼>

x < 6x²
6x²-x > 0
x(6x-1) > 0
x = ⅙
x∈ (0,⅙)

(-x+5)(x+2) > 0

-x+5 = 0  x = 5

x+2 = 0  x = -2

x=5 ∨ x=-2
x∈ (-2,5)

(1+x)(3-2x) ≤ 0

1+x = 0 x=-1

3-2x = 0 2x=3 x= 1,5

x=-1 ∨ x=1,5
x∈(-∞,-1>∨<1,5,+∞)

x²<1

x²-1 < 0 

(x+1)(x-1) < 0
x=1 ∨ x=-1
x∈(-1,1)

3x²-1>2½ x
3x²-2½x-1>0
Δ=b²-4ac

Δ = (-5/2)²-4*3*(-1) = 25/4+12 = 25/4+48/4 = 73/4

√Δ = √73 / 2

x₁ = 5/2 - √73 / 2 = 5-√73  /2

x₂ = 5+√73 /2

x (-∞, 5-√73  /2  v 5+√73 / 2 , +∞)


x-7 ≥ 5x
-4x-7 ≥ 0
-4 x ≥ 7 /:(-4)
x ≥ -7/4
 

½ x²-x ≥ 1
½x²-x-1≥0
Δ = (-1)²-4*1/2*(-1) = 1+2 = 3

√Δ = √3

x₁ = 1-√3

x² = 1+√3

(-∞,1-√3> v <1+√3,+∞)

4x²+49≤28x
4x²-28x+49≤0
(2x-7) ≤ 0

2x ≤ 7 

x ≤ 3,5