Մեկ անհայտով, երկրորդ աստիճանի անհավասարում. 15.11.2022

200.

ա) 2x – 3 ≤ 0
2x – 3 = 0
2x = 3
x = 3/2 = 1,5
x ∈ (-∞; 1,5]

բ) 4x – 3 ≥ 0
4x – 3 = 0
4x = 3
x = 3/4 = 0,75
x ∈ [0,75; +∞)

գ) 5x – 8 ≥ 3x – 1
5x – 3x – 8 + 1 ≥ 0
2x – 7 ≥ 0
2x – 7 = 0
2x = 7
x = 7/2 = 3,5
x ∈ [3,5; +∞)

դ) 2x – 4 ≥ 4x – 3
2x – 4x – 4 + 3 ≥ 0
-2x – 1 ≥ 0
-2x – 1 = 0
2x = -1
x = -1/2 = -0,5
x ∈ (-∞; -0,5]

201.

ա) x2 – 12x + 32 ≤ 0
x2 – 12x + 32 = 0
{x1 ⋅ x2 = 32
{x1 + x2 = 12
x1 = 8
x2 = 4
x ∈ [4; +∞)

բ) x2 + 8x + 12 ≤ 0
x2 + 8x + 12 = 0
{x1 ⋅ x2 = 12
{x1 + x2 = -8
x1 = -2
x2 = -6
x ∈ [-6; -2]

գ) 2x2 + x – 7 ≥ 0
2x2 + x – 7 = 0
D = 1 + 56 = 57
x = (-1 ± 57)/4

դ) 3x2 – 5x – 1 ≤ 0
3x2 – 5x – 1 = 0
D = 25 + 12 = 37
x = (-1 ± 37)/4

202.

ա) -2x2 + 2x – 1 ≥ 0
-2x2 + 2x – 1 = 0
D = 4 – 8 = -4
Արմատներ չունի

բ) -x2 + 4x – 4 ≤ 0
-x2 + 4x – 4 = 0
D = 16 – 16 = 0
x = (-1 ± 0)/-2 = 0,5
x ∈ (-∞; +∞)

գ) 3x2 + 18x + 27 ≤ 0
3x2 + 18x + 27 = 0
x2 + 6x + 9 = 0
{x1 ⋅ x2 = 9
{x1 + x2 = -6
x = -3

դ) 2x2 – 20x + 50 ≥ 0
2x2 – 20x + 50 = 0
x2 – 10x + 25 = 0
{x1 ⋅ x2 = 25
{x1 + x2 = 10
x = 5
x ∈ (-∞; +∞)

203.

ա) x2 – 3x + 5 ≥ 0
x2 – 3x + 5 = 0
D = 9 – 20 = -11
Արմատներ չունի

բ) x2 + 7x + 10 ≤ 0
x2 + 7x + 10 = 0
D = 49 – 40 = 9
x1 = (-7 + 3)/2 = -2
x2 = (-7 – 3)/2 = -5
x ∈ [-5; -2]

գ) 8x2 – x + 1 ≤ 0
8x2 – x + 1 = 0
D = 1 – 32 = -31
Արմատներ չունի

դ) 4x2 – 5x + 6 ≥ 0
4x2 – 5x + 6 = 0
D = 25 – 96 = -71
Արմատներ չունի

204.

ա) (x2 – 1)(x + 3) ≥ 0
(x2 – 1)(x + 3) = 0
[x2 – 1 = 0 | x2 = 1 | x = ±1
[x + 3 = 0 | x = -3
x ∈ [-3; -1] ∪ [1; +∞)

բ) (7 – x)(4 – x2) ≤ 0
(7 – x)(4 – x2) = 0
[7 – x = 0 | x = 7
[4 – x2 = 0 | x2 = -4

գ) (12 – 5x)(x2 – 4x + 4) ≤ 0
(12 – 5x)(x2 – 4x + 4) = 0
[12 – 5x = 0 | 5x = 12 | x = 12/5 = 2,4
[x2 – 4x + 4 = 0 | x = 2
x ∈ [2; 24]

դ) (x2 – 5x + 6)(x – 3) ≤ 0
(x2 – 5x + 6)(x – 3) = 0

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