Question

помогите решить систему неравенств..

 

Answer 1

ОДЗ: х > 0 и х не равен 1/3

1 неравенство:

2*9*3^x + 27/3^x <= 87 замена а = 3^x

18a^2 - 87a + 27 <= 0

D = 87*87 - 4*18*27 = 3*29*3*29 - 4*2*9*27 = 9*(29*29 - 8*27) = 9*625 = 75*75

a1 = (87-75)/36 = 12/36 = 1/3

a1 = (87+75)/36 = 162/36 = 9/2 = 4.5

решение неравенства 1/3 <= а <= 4.5

1/3 <= 3^x <= 4.5

3^(-1) <= 3^x <= 3^(log(3)(4.5))

-1 <= x <= log(3)(4.5)

2 неравенство:

-3*log(3x)(3) * log(3)(27x) + 9 >= 0

-3/log(3)(3x) * log(3)(9*3x) + 9 >= 0

-3(log(3)(9)+log(3)(3x) / log(3)(3x) + 9 >= 0

-3*2/log(3)(3x) -3 + 9 >= 0

-6/log(3)(3x) + 6 >= 0 разделим на -6

1 / log(3)(3x) - 1 <= 0

1 / log(3)(3x) <= 1

log(3)(3x) >= 1

3x >= 3

x >= 1

log(3)(4.5) = log(3)(9*0.5) = 2+log(3)(0.5) > 1

т.к. log(3)(1/3) = -1, 0.5 > 1/3 => log(3)(0.5) > log(3)(1/3) основание 3>1 ---функция возрастающая => log(3)(0.5) > -1 => 2+log(3)(0.5) > 2-1

Ответ: 1 <= x <= log(3)(4.5)