Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 281

1) 25 – х² ≥ 0
25 – х² = 0
х² = 25
х = ± $$\sqrt{25}$$
х = ± 5

2) 9х – х² – 14 ≥ 0

D = 9² – 4 · (–1) · (–14) = 81 – 56 = 25

х₁ = $$\frac{-9 + \sqrt{25}}{2 \cdot (-1)}$$ = –9 + 5/–2 = –4/–2 = 2

х₂ = $$\frac{-9 - \sqrt{25}}{2 \cdot (-1)}$$ = –9 – 5/–2 = –14/–2 = 7

х ∈ [2; 5]      =>      х ∈ {2; 3; 4; 5}

Ответ: х ∈ {2; 3; 4; 5}

1) 8х – х² – 12 ≥ 0

D = 8² – 4 · (–1) · (–12) = 64 – 48 = 16

х₁ = $$\frac{-8 + \sqrt{16}}{2 \cdot (-1)}$$ = –8 + 4/–2 = –4/–2 = 2

х₂ = $$\frac{-8 - \sqrt{16}}{2 \cdot (-1)}$$ = –8 – 4/–2 = –12/–2 = 6

х ∈ [2; 4]      =>      х ∈ {2; 3; 4}

Ответ: х ∈ {2; 3; 4}