a-ի տեղը y եմ օգտագործել որպեսզի չխառնվի։

5․ (y² – 9)x² – (2y + 6)x + 1 = 0

D < 0
a = (y² – 9)
b = -(2y + 6)
c = 1

D = b² – 4ac = (-(2y + 6))² – 4 · (y² – 9) = 4y² + 24y + 36 – 4y² + 36 = 24y + 72
D = 0
24y + 72 = 0
24y = -72
y = -3

D < 0
y < -3

Պատ․ y ∈ (-∞; -3)

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6. (y + 2)²x² – (y + 2)x² – 1 = 0

D < 0
a = (y + 2)²
b = -(y + 2)
c = -1

D = b² – 4ac = (-(y + 2))² – 4 · (y + 2)² · -1 = (y + 2)² + 4(y + 2)² = 5(y + 2)²
D = 0
5(y + 2)² = 0
5(y + 2) = 0
y + 2 = 0
y = -2

D < 0
y < -2

Պատ․ y ∈ (-∞; -2)

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8. (y – 1)x² – x(y – 1) + 1 = 0

D ≥ 0
a = y – 1
b = -(y – 1)
c = 1

D = b² – 4ac = (-(y – 1))² – 4 · (y – 1) = y² – 2y + 1 – 4y + 4 = y² – 6y + 5

D = 0
y² – 6y + 5 = 0

y₁y₂ = c/a = 5
y₁ + y₂ = -b/a = 6
y₁ = 1
y₂ = 5

D ≥ 0
y² – 6y + 5 ≥ 0
a > 0
y ∈ (-∞; 1] ∪ [5; +∞)

a ≠ 0
y – 1 ≠ 0
y ≠ 1

Պատ․ y ∈ (-∞; 1) ∪ [5; +∞)

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9. (y² – 4)x² + (y + 2)x + 2 = 0

D ≥ 0
a = y² – 4
b = y + 2
c = 2

D = b² – 4ac = (y + 2)² – 4 · (y² – 4) · 2  = (y + 2)² – 8(y – 2)(y + 2) = (y + 2)(y + 2 – 8y + 16 = (y + 2)(18 – 7y)
D = 0
(y + 2)(18 – 7y) = 0
⌈y + 2 = 0 ⇒ y = -2
⌊18 – 7y = 0 ⇒ y = ¹⁸/₇

D ≥ 0 
(y + 2)(18 – 7y) ≥ 0
a < 0
y ∈ [-2; ¹⁸/₇]
a ≠ 0
y² – 4 ≠ 0
y² ≠ 4
y ≠ ±2

Պատ․ y ∈ (-2; ¹⁸/₇] \ 2

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12. yx² – 2(y – 1)x + y + 1 = 0

D > 0
a = y
b = -2(y – 1)
c = y + 1

D = b² – 4ac = (-2(y – 1))² – 4 · y · (y + 1) = (-2y + 2)² – 4y² + 4y = 4y² – 8y + 4 – 4y² + 4y = 4 – 4y
D = 0
4 – 4y = 0
4y = 4
y = 1

D > 0
y > 1

a ≠ 0
y ≠ 0

Պատ․ y ∈ (1; +∞)

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16. x² + 2(y – 1)x + y² – y – 2

D > 0 
c/a > 0

a = 1
b = 2(y – 1)
c = y² – y – 2

D = b² – 4ac = (2(y – 1))² – 4(y² – y – 2) = 4(y² – 2y + 1) – 4(y² – y – 2) = 4(y² – 2y + 1 – y² + y + 2) = 4(3 – y) = 12 – 4y

D = 0
12 – 4y = 0
12 = 4y
y = 3
D > 0
y > 3

c/a = 0
y² – y – 2 = 0
a = 1, b = -1, c = -2
x₁ + x₂ = -b/a = 1
x₁x₂ = c/a = -2

x₁ = -2
x₂ = 1

a > 0
y² – y – 2 > 0
y ∈ (-∞; -2) ∪ (1; +∞)
y ∈ ((-∞; -2) ∪ (1; +∞)) ∩ (3; +∞)

Պատ․ y > 3