1) (x+1)² (x² + 2x) = 12
(x² + 2x + 1) (x² + 2x) = 12
x² + 2x = t
(t+1) t = 12
t² + t - 12 = 0
D = 1+48 = 49
t₁ = 3
t₂ = -4
x² + 2x = 3
x² + 2x - 3 = 0
x₁ = 1
x₂ = -3
x² + 2x = -4
x² + 2x + 4 =0
D = 4-16 <0<br>нет корней
Ответ: 1; -3
2) x² +3x = t
(t+1) (t+3) = -1
t² + 3t + t + 3 = -1
t² + 4t + 4 = 0
D₁ = 4-4 = 0
t = -2
x² +3x = -2
x² + 3x + 2=0
x₁ = -1
x₂ = -2
Ответ: -1; -2
3) x² - 4x = t
(t+1) (t+2) = 12
t² + 2t + t + 2 = 12
t² + 3t - 10 = 0
D = 9 + 40 = 49
t₁ = (-3+7) / 2 = 2
t₂ = (-3-7) / 2 = -5
x² - 4x = 2
x² - 4x - 2 = 0
D₁ = 4+2 = 6
x₁ = 2+√6
x₂ = 2-√6
x² - 4x = -5
x² - 4x + 5 = 0
D₁ = 4 - 5 < 0
нет корней
Ответ: 2+√6 ; 2-√6