X²-17x+72>0
x²-17x+72=0 D=1
x₁=8 x₂=9 ⇒
(x-8)(x-9)>0
-∞_______+_______8______-_______9_______+_______+∞
x∈(-∞;8)U(9;+∞).
Пусть t = (x-2)^2 ≥0
Тогда получим уравнение: t^2-t-6=0
Корни этого уравнения t = 3, t=-2
Последний корень не подходит, т.к. t>0, a -2<0
Получаем
(x-2)^2 = 3
x^2 - 4x +1 =0
D = 16-4= 12 = (2√3)²
x = 2-√3 и х=2+√3
Log8 32-log8 1/2= <span>log8 ( 32 / 1/2)= log8 64 = 2</span>
Решение
1) b₁ = 3
Sn = 6
Sn =b₁ / (1 - q)
6*(1 - q) = 3
6 - 6q = 3
6q = 3
q = 1/2
b₂ = b₁ * q
b₂ = 3 * (1/2)
b₂ = 1,5
2) b₇ - b₄ = 168; q = 2
b₇ = b₁ * q⁶
b₄ = b₁*q³
b₁ * q⁶ - b₁*q³ = 168
b₁*(q⁶ - q³) = 168
b₁ =168 / (2⁶ - 2³)
b₁ = 168 / (64 - 8)
b₁ = 168 / 56
b₁ = 3
3) bn = b₁ * q^(n - 1)
q = 16/8 = 2
512 = 8 * [2^(n - 1)]
64 = 2^(n - 1)
2⁶ = 2^(n - 1)
n - 1 = 6
n = 7
4) b₄ - b ₁ = 52
b₁ + b₂ + b₃ = 26
b₁*q³ - b₁ = 52
b₁ + b₁q + b₁q² = 26
b₁(q² - 1) = 52, q³ - 1 ≠ 0, q ≠ 1
b₁( 1 + q + q²) = 26
b₁(q - 1)*(q² + q + 1) = 52
26*(q - 1) = 52
26q = 26 + 52
26q = 78
q = 3
b₁(q³ - 1) = 52
b₁ = 52/(q³ - 1)
b₁ = 52/(26)
b₁ = 2
S₆ = [b₁(1 - q⁶)] / (1 - q)
S₆ = [2*(1 - 3⁶) / (1 - 3)]
S₆ = {2*(1 - 729) / (- 2)]
S₆ = 728