Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 339

а) х² – 5х – 50 < 0

х² – 5х – 50 = 0

D = (–5)² – 4 · 1 · 50 = 25 + 200 = 225

x₁ = $$\frac{5 + \sqrt{225}}{2 \cdot 1}$$ = 5 + 15/2 = 20/2 = 10

x₂ = $$\frac{5 - \sqrt{225}}{2 \cdot 1}$$ = 5 – 15/2 = –10/2 = –5

Ответ: х ∈ (–5; 10)

б) –m² – 8m + 9 ≥ 0

m² + 8m – 9 ≥ 0 | · (–1)
m² + 8m – 9 = 0

D = 8² – 4 · 1 · ( –9) = 64 + 36 = 100

x₁ = $$\frac{-8 + \sqrt{100}}{2 \cdot 1}$$ = –8 + 10/2 = 2/2 = 1

x₂ = $$\frac{-8 - \sqrt{100}}{2 \cdot 1}$$ = –8 – 10/2 = –18/2 = –9

Ответ: m ∈ [–9; 1]

в) 3y² + 4y – 4 > 0

3у² + 4у – 4 = 0

D = 4² – 4 · 3 · (–4) = 16 + 48 = 64

y₁ = $$\frac{-4 + \sqrt{64}}{2 \cdot 3}$$ = –4 + 8/6 = 4/6 = 2/3

y₂ = $$\frac{-4 - \sqrt{64}}{2 \cdot 3}$$ = –4 – 8/6 = –12/6 = –2

Ответ: y ∈ (– ∞; –2) ∪ (2/3; + ∞)

г) 8p² + 2p ≥ 21

8р² + 2р – 21 ≥ 0
8р² + 2р – 21 = 0

D = 2² – 4 · 8 · (–21) = 4 + 672 = 676

p₁ = $$\frac{-2 + \sqrt{676}}{2 \cdot 8}$$ = –2 + 26/16 = 24/16 = 3/2 = 1,5

p₂ = $$\frac{-2 - \sqrt{676}}{2 \cdot 8}$$ = –2 – 26/16 = –28/16 = –7/4 = –1,75

Ответ: p ∈ (– ∞; –1.75] ∪ [1,5; + ∞)

д) 12x – 9 ≤ 4x²

12х – 9 – 4х² ≤ 0 | · (–1)
4х² – 12х + 9 ≤ 0
4х² – 12х + 9 = 0
(2х – 3)² = 0
2х – 3 = 0
2х = 3
х = 1,5

Ответ: x ∈ (– ∞; + ∞)

е) –9x < 1 – 6x

–9х² + 6х – 1 < 0 | · (–1)
9х² – 6х + 1 > 0
9х² – 6х + 1 = 0
(3х – 1)² = 0
3х – 1 = 0
3х = 1
х = 1/3

Ответ: x ∈ (– ∞; 1/3) ∪ (1/3; + ∞)