Первое: (4/9)^x<=sqrt(2/3)
2x>=1/2
x>=1/4
Второе: замена t=log3(x), t!=0
(t-7)/(1/t-3)<=2
(7t-t^2)/(3t-1)<=2
(t^2-t-2)/(3t-1)>=0
(t-2)(t+1)/(3t-1)>=0
t in [-1, 0) U (0, 1/3)U[2,+infty)
x in [1/3, 1) U (1, 3^(1/3)) U [9, +infty)
Пересекаем с решением первого, в итоге имеем [1/3, 1) U (1, 3^(1/3)) U [9, +infty)
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Tga=1/3
cos²a=1:(1+tg²a)=1:(1+1/9)=1:10/9=9/10
2cos²a+1=2*9/10+1=1,8+1=2,8
X1=5
25+5a+20=0
5a=-45
a=-9
x1*x2=20
x2=20:5
x2=4