ОДЗ: х > 0 и х не равен 1/3
1 неравенство:
2*9*3^x + 27/3^x <= 87 замена а = 3^x</p>
18a^2 - 87a + 27 <= 0</span>
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</p>
1/3 <= 3^x <= 4.5</p>
3^(-1) <= 3^x <= 3^(log(3)(4.5))</p>
-1 <= x <= log(3)(4.5)</u>
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</p>
1 / log(3)(3x) <= 1</p>
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)