Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 266

а) 2х² + 13х − 7 > 0

D = 13²– 4 · 2 · (–7) = 169 + 56 = 225

х₁ = $$\frac{-13 + \sqrt{225}}{2 \cdot 2}$$ = –13 + 15/4 = 2/4 = 0,5

х₂ = $$\frac{-13 - \sqrt{225}}{2 \cdot 2}$$ = –13 – 15/4 = –28/4 = –7

Ответ: х ∈ (– ∞; –7] ∪ [0,5; + ∞)

б) −9х² + 12х − 4 < 0

−(3х – 2)² = 0
(3х – 2)² = 0
3х – 2 = 0
3х = 2
х = 2/3

Ответ: х ∈ (– ∞; 2/3) ∪ (2/3; + ∞)

в) 6х² − 13х + 5 ≤ 0

D = (–13)² – 4 · 6 · 5 = 169 – 120 = 49

х₁ = $$\frac{13 + \sqrt{49}}{2 \cdot 6}$$ = 13 + 7/12 = 20/12 = 5/3 = 12/3

х₂ = $$\frac{13 - \sqrt{49}}{2 \cdot 6}$$ = 13 – 7/12 = 6/12 = 1/2

Ответ: х ∈ [1/2; 11/2]

г) −2х² − 5х + 18 ≤ 0

D = (–5)² – 4 · (–2) · 18 = 25 + 144 = 169

х₁ = $$\frac{5 + \sqrt{169}}{2 \cdot (-2)}$$ = 5 + 13/–4 = 18/–4 = –4,5

х₂ = $$\frac{5 - \sqrt{169}}{2 \cdot (-2)}$$ = 5 – 13/–4 = –8/–4 = 2

Ответ: х ∈ (– ∞; –4,5] ∪ [2; + ∞)

д) 3х² − 2х > 0

3х² – 2х = 0
х(3х – 2) = 0
х = 0  или  3х – 2 = 0
                    3х = 2
                    х = 2/3

Ответ: х ∈ (– ∞; 0] ∪ [2/3; + ∞)

е) 8 − х² < 0

8 – х² = 0
х² = 8
х = ± $$\sqrt{8}$$
х = ± 2$$\sqrt{2}$$

Ответ: х ∈ (– ∞; –2$$\sqrt{2}$$] ∪ [2$$\sqrt{2}$$; + ∞)