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IFender
@IFender
September 2021
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РЕШИТЕ ПЕРВЫЕ 3 ПРИМЕРА
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nKrynka
Решение
1) 2^(3x/2) * 2^(x - 5) = 1
2^(3x/2 + x - 5) = 2^0
3x/2 + x - 5 = 0
3x + 2x - 10 = 0
5x = 10
x = 2
x₀ = 2
(0,2)^(-x₀) = (1/5)⁻² = 5² = 25
2) 2^(x - 2) + 5/[2^(3 - x)] = 28
2^x * 2⁻² + (5*2^x) /2³ = 28
2^x * (1/4 + 5/8) = 28
2^x * (7/8) = 28
2^x = 32
2^x = 2⁵
x = 5
3) 3^(4x - 1) + 3^(2x) - 6 = 0
(1/3)*3^(4x) + 3^(2x) - 6 = 0 умножаем обе части уравнения на 3
3^(4x) + 3*3^(2x) - 18 = 0
Пусть 3^(2x) = t, t > 0
t² + 3t - 18 = 0
t₁ = - 6 < 0, не удовлетворяет условию t > 0
t₂ = 3
3^(2x) = 3
3^(2x) = 3¹
2x = 1
x = 1/2
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Answers & Comments
1) 2^(3x/2) * 2^(x - 5) = 1
2^(3x/2 + x - 5) = 2^0
3x/2 + x - 5 = 0
3x + 2x - 10 = 0
5x = 10
x = 2
x₀ = 2
(0,2)^(-x₀) = (1/5)⁻² = 5² = 25
2) 2^(x - 2) + 5/[2^(3 - x)] = 28
2^x * 2⁻² + (5*2^x) /2³ = 28
2^x * (1/4 + 5/8) = 28
2^x * (7/8) = 28
2^x = 32
2^x = 2⁵
x = 5
3) 3^(4x - 1) + 3^(2x) - 6 = 0
(1/3)*3^(4x) + 3^(2x) - 6 = 0 умножаем обе части уравнения на 3
3^(4x) + 3*3^(2x) - 18 = 0
Пусть 3^(2x) = t, t > 0
t² + 3t - 18 = 0
t₁ = - 6 < 0, не удовлетворяет условию t > 0
t₂ = 3
3^(2x) = 3
3^(2x) = 3¹
2x = 1
x = 1/2