Алгебра 9 класс учебник Макарычев, Миндюк ответы – номер 221

а) (2х² + 3)² – 12(2х² + 3) + 11 = 0

Замена: 2х² + 3 = t

t² – 12t + 11 = 0
D = (–12)² – 4 · 1 · 11 = 144 – 44 = 100

При t = 1                           При t = 11
2х² + 3 = 1                        2х² + 3 = 11
2х² = 1 – 3                        2х² = 11 – 3
2х² = –2                            2х² = 8
х² = –1 – корней нет      х² = 4
                                           х = ± $$\sqrt{4}$$
                                           х = ± 2
Ответ: х = ± 2

б) (t² – 2t)² – 3 = 2(t² – 2t)

Замена: t² – 2t = х

х² – 3 = 2х
х² – 2х – 3 = 0
D = (–2)² – 4 · 1 · (–3) = 4 + 12 = 16

При х = 3
t² – 2t = 3
t² – 2t – 3 = 0
D = (–2)² – 4 · 1 · ( –3) = 4 + 12 = 16

При х = –1
t² – 2t = –1
t² – 2t + 1 = 0
(t – 1)² = 0
t – 1 = 0
t = 1
Ответ: t = –1; t = 1; t = 3

в) (x² + x – 1)(x² + x + 2) = 40

Замена: х² + х = t

(t – 1)(t + 2) = 40
t² + 2t – t – 2 – 40 = 0
t² + t – 42 = 0
D = (–1)² – 4 · 1 · (–42) = 1 + 168 = 169

При t = 6
х² + х = 6
х² + х – 6 = 0
D = 1² – 4 · 1 · (–6) = 1 + 24 = 25

При t = –7
х² + х = –7
х² + х + 7 = 0
D = 1² – 4 · 1 · 7 = 1 – 28 = –27 < 0 – корней нет
Ответ: х = –3; х = 2

г) (2x² + x – 1)(2x² + x – 4) + 2 = 0

Замена: 2х² + х = t

(t – 1)(t – 4) + 2 = 0
t² – 4t – t + 4 + 2 = 0
t² – 5t + 6 = 0
D = (–5)² – 4 · 1 · 6 = 25 – 24 = 1
t₁ = 5 + 1/2 · 1 = 6/2 = 3
t₂ = 5 – 1/2 · 1 = 4/2 = 2

При t = 3
2х² + х = 3
2х² + х – 3 = 0
D = 1² – 4 · 2 · (–3) = 1 + 24 = 25

При t = 2
2х² + х = 2
2х² + х – 2 = 0
D = 1² – 4 · 2 · (–2) = 1 + 16 = 17

Ответ: х = –1,5; х = 1; х =