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Masha181198
@Masha181198
July 2021
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решите первые 4 номера пожалуйста, очень срочно нужно
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nKrynka
Решение
1) f(x) = x³ - 3x² - 24x + 17
f`(x) ≥ 0
f`(x) = 3x² - 6x - 24
x² - 2x - 8 ≥ 0
x₁ = - 2
x₂ = 4
+ - +
------------------------------------------------------------------------->
- 2 4 x
x ∈ ( - ∞; - 2] [4 ; + ∞)
2) 2tg(5π/8) / (1 - tg²(5π/8) = tg2*(5π/8) = tg(5π/4)
3) √(3x² + 5x - 7) = √(5 - 4x)
[√(3x² + 5x - 7)]² = [√(5 - 4x)]²
3x² + 5x - 7 = 5 - 4x
3x² + 9x - 12 = 0
x² + 3x - 4 = 0
x₁ = - 4
x₂ = 1
Проверка
x = - 4
√[3*(- 4)² + 5*(- 4) - 7)] = √(48 - 20 - 7) = √21
√[5 - 4*(- 4)] = √(5 + 16) = √21
√21 = √21 верно
x = 1
√(3*1² + 5*1 - 7) = = √(3 + 5 - 7 ) = √1 = 1
√(5 - 4*1) = √1 = 1
1 = 1 верно
Ответ: x₁ = - 4; x₂ = 1
4) 7 * 81³/⁴ - 9*32³/⁵ + 5 * 49¹/² = 7 * (3⁴)³/⁴ - 9 * (2⁵)³/⁵ + 5 *( 7²)¹/² =
= 7 * 3³ - 9 * 2³ + 5 * 7 = 7 * 27 - 9 * 8 + 5 * 7 = 189 - 72 + 35 = 152
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Masha181198
спасибо большое
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Answers & Comments
1) f(x) = x³ - 3x² - 24x + 17
f`(x) ≥ 0
f`(x) = 3x² - 6x - 24
x² - 2x - 8 ≥ 0
x₁ = - 2
x₂ = 4
+ - +
------------------------------------------------------------------------->
- 2 4 x
x ∈ ( - ∞; - 2] [4 ; + ∞)
2) 2tg(5π/8) / (1 - tg²(5π/8) = tg2*(5π/8) = tg(5π/4)
3) √(3x² + 5x - 7) = √(5 - 4x)
[√(3x² + 5x - 7)]² = [√(5 - 4x)]²
3x² + 5x - 7 = 5 - 4x
3x² + 9x - 12 = 0
x² + 3x - 4 = 0
x₁ = - 4
x₂ = 1
Проверка
x = - 4
√[3*(- 4)² + 5*(- 4) - 7)] = √(48 - 20 - 7) = √21
√[5 - 4*(- 4)] = √(5 + 16) = √21
√21 = √21 верно
x = 1
√(3*1² + 5*1 - 7) = = √(3 + 5 - 7 ) = √1 = 1
√(5 - 4*1) = √1 = 1
1 = 1 верно
Ответ: x₁ = - 4; x₂ = 1
4) 7 * 81³/⁴ - 9*32³/⁵ + 5 * 49¹/² = 7 * (3⁴)³/⁴ - 9 * (2⁵)³/⁵ + 5 *( 7²)¹/² =
= 7 * 3³ - 9 * 2³ + 5 * 7 = 7 * 27 - 9 * 8 + 5 * 7 = 189 - 72 + 35 = 152