Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 280

1) х² – 2х – 8 < 0

D = (–2)² – 4 · 1 · (–8) = 4 + 32 = 36

х₁ = $$\frac{2 + \sqrt{36}}{2}$$ = 2 + 6/2 = 8/2 = 4

х₂ = $$\frac{2 - \sqrt{36}}{2}$$ = 2 – 6/2 = –4/2 = –2

2) x² – 9 < 0
х² – 9 = 0
х² = 9
х = ± $$\sqrt{9}$$
х = ± 3

Ответ: х ∈ (–2; 3)

1) 2х² – 13х + 6 < 0

D = (–13)² – 4 · 2 · 6 = 169 – 48 = 121

х₁ = $$\frac{13 + \sqrt{121}}{2 \cdot 2}$$ = 13 + 11/4 = 24/4 = 6

х₂ = $$\frac{13 - \sqrt{121}}{2 \cdot 2}$$ = 13 – 11/4 = 2/4 = 0,5

2) x² – 4x > 0
x² – 4x = 0
х(х – 4) = 0
х = 0  или  х – 4 = 0
                    х = 4

Ответ: х ∈ (4; 6)

1) х² – 6х – 16 > 0

D = (–6)² – 4 · 1 · (–16) = 36 + 64 = 100

х₁ = $$\frac{6 + \sqrt{100}}{2 \cdot 1}$$ = 6 + 10/2 = 16/2 = 8

х₂ = $$\frac{6 - \sqrt{100}}{2 \cdot 1}$$ = 6 – 10/2 = –4/2 = –2

2) x² + 2x – 120 < 0

D = 2² – 4 · 1 · (–120) = 4 + 480 = 484

х₁ = $$\frac{-2 + \sqrt{484}}{2 \cdot 1}$$ = –2 + 22/2 = 20/2 = 10

х₂ = $$\frac{-2 - \sqrt{484}}{2 \cdot 1}$$ = –2 – 22/2 = –24/2 = –12

Ответ: х ∈ (–12; –2) ∪ (8; 10)

1) 3х² + х – 2 ≤ 0

D = 1² – 4 · 3 · (–2) = 1 + 24 = 25

х₁ = $$\frac{-1 + \sqrt{25}}{2 \cdot 3}$$ = –1 + 5/6 = 4/6 = 2/3

х₂ = $$\frac{-1 - \sqrt{25}}{2 \cdot 3}$$ = –1 – 5/6 = –6/6 = –1

2) х² + 4х – 12 ≤ 0

D = 4² – 4 · 1 · (–12) = 16 + 48 = 64

х₁ = $$\frac{-4 + \sqrt{64}}{2 \cdot 1}$$ = –4 + 8/2 = 4/2 = 2

х₂ = $$\frac{-4 - \sqrt{64}}{2 \cdot 1}$$ = –4 – 8/2 = –12/2 = –6

Ответ: х ∈ [1; 2/3]

1) 2х² + 4х + 15 ≤ 0

D = 4² – 4 · 2 · 15 = 16 – 120 = –104 < 0

х – любое число

2) х² – 9х + 8 ≤ 0

D = (–9)² – 4 · 1 · 8 = 81 – 32 = 49

х₁ = $$\frac{9 + \sqrt{49}}{2}$$ = 9 + 7/2 = 16/2 = 8

х₂ = $$\frac{9 - \sqrt{49}}{2}$$ = 9 – 7/2 = 2/2 = 1

Ответ: х ∈ [1; 8]

1) 2х² + 5х – 3 < 0

D = 5² – 4 · 2 · (–3) = 25 + 24 = 49

х₁ = $$\frac{-5 + \sqrt{49}}{2 \cdot 2}$$ = –5 + 7/4 = 2/4 = 0,5

х₂ = $$\frac{-5 - \sqrt{49}}{2 \cdot 2}$$ = –5 – 7/4 = –12/4 = –3

2) 3х² + х + 11 < 0

D = 1² – 4 · 3 · 11 = 1 – 132 = - 131 < 0

Следовательно, х ∈ ∅ и система не имеет решений.

Ответ: х ∈ ∅