Log₃ (x² - 4x + 4) = 2
x² - 4x + 4 = 9
x² - 4x - 5 = 0
По теореме Виета:
x₁ = -1
x₂ = 5
3log₄ x = log₄ 12,5 + log₄ 64
log₄ x³ = log₄ 800
x³ = 800
x = 2
![\sqrt[3]{100}](https://tex.z-dn.net/?f=+%5Csqrt%5B3%5D%7B100%7D+)
2log₃ (x-2) - log₃ (x+1) = 1
log₃ (x-2)² - log₃ (x+1) = 1
log₃ (x-2)² = log₃ 3 + log₃ (x+1)
log₃ (x-2)² = log₃ 3(x+1)
x² - 4x + 4 = 3x + 3
x² - 7x + 1 = 0
D = (-7)² - 4 = 45
x₁ =
![\frac{7- \sqrt{45} }{2}](https://tex.z-dn.net/?f=+%5Cfrac%7B7-+%5Csqrt%7B45%7D+%7D%7B2%7D+)
x₂ =
![\frac{7+ \sqrt{45} }{2}](https://tex.z-dn.net/?f=+%5Cfrac%7B7%2B+%5Csqrt%7B45%7D+%7D%7B2%7D+)
<span>log₄ (x-4) + log₄ (x+4) = log₄ (3x+2)
</span>log₄ (x-4)(x+4) = log₄ (3x+2)
x² - 16 = 3x+2
x² - 3x - 18 = 0
По теореме Виета:
x₁ = -3
x₂ = 6
У меня 2 варианта
( не очень понятно условие задания)
![3 \sqrt{x} -12=0 \\ 3 \sqrt{x} =12 \\ \sqrt{x} =4 \\ x=4 ^{2} \\ x=16 \\ \\ 3 \sqrt{x-12} =0 \\ \sqrt{x-12} =0 \\ x-12=0 \\ x=12](https://tex.z-dn.net/?f=3+%5Csqrt%7Bx%7D+-12%3D0+%5C%5C+3+%5Csqrt%7Bx%7D+%3D12+%5C%5C++%5Csqrt%7Bx%7D+%3D4+%5C%5C+x%3D4+%5E%7B2%7D++%5C%5C+x%3D16+%5C%5C++%5C%5C+3+%5Csqrt%7Bx-12%7D+%3D0+%5C%5C++%5Csqrt%7Bx-12%7D+%3D0+%5C%5C+x-12%3D0+%5C%5C+x%3D12)
(1-sqrt(2)+1+sqrt(2))/((1-sqrt(2)(1+sqrt(2))=-6.24