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Алексей334
@Алексей334
July 2022
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(3x^2-4)^2-4(bx^2-4)-5=0 сколько будет
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Answers & Comments
meripoppins60
(3x² - 4)² - 4(3x²
- 4) - 5 = 0
Ввести новую переменную
t =
3x² - 4
t² - 4t - 5 = 0
а = 1; b = -4; c = -5
D = b² - 4ac = (-4)² - 4 * 1 * (-5) = 16 + 20 = 36
t
₁
=
- b + √D
=
- ( -4) + √36
=
4 + 6
= 5
2a 2 * 1 2
t
₂
=
- b - √D
=
- ( - 4) - √36
=
4 - 6
= -1
2a 2 * 1 2
При t₁ = 5,
t = 3x² - 4
5 =
3x² - 4
3x² = 9
x² = 3
x₁ = -√3, x₂ = √3
При t₂ = -1,
t = 3x² - 4
-1 = 3x² - 4
3x² = 3
x² = 1
x₁ = -1, x₂ = 1
Ответ:
-√3, -1, 1, √3
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Answers & Comments
Ввести новую переменную
t = 3x² - 4
t² - 4t - 5 = 0
а = 1; b = -4; c = -5
D = b² - 4ac = (-4)² - 4 * 1 * (-5) = 16 + 20 = 36
t₁ = - b + √D = - ( -4) + √36 = 4 + 6 = 5
2a 2 * 1 2
t₂ = - b - √D = - ( - 4) - √36 = 4 - 6 = -1
2a 2 * 1 2
При t₁ = 5,
t = 3x² - 4
5 = 3x² - 4
3x² = 9
x² = 3
x₁ = -√3, x₂ = √3
При t₂ = -1,
t = 3x² - 4
-1 = 3x² - 4
3x² = 3
x² = 1
x₁ = -1, x₂ = 1
Ответ: -√3, -1, 1, √3