Log(3+x) числа (9-x^2)-1/16log^2(x+3) числа (x-3)^2>=2

Question

Log(3+x) числа (9-x^2)-1/16log^2(x+3) числа (x-3)^2>=2

Answer 1

Log(x+3)(9-x^2)-(log(x+3)(x-3)^2)^2/16 > =2 ОДЗ -3 < x < 3 
log(x+3)(x+3)+log(x+3)(3-x)-(log(x+3)(3-x)+log(x+3)(3-x))^2/16 > =2 
1+log(x+3)(3-x)-(log(x+3)(3-x)+log(x+3)(3-x))^2/16 > =2 
log(x+3)(3-x)-(log(x+3)(3-x)+log(x+3)(3-x))^2/16 > =1 
y=Log(x+3)(3-x) 
y-2y^2/16 > =1 
4y-y^2 > =4 
y^2-4y+4 < =0 
y=2 
log(x+3)(3-x)=2 
3-x=(x+3)^2 
x^2+7x+6=0 
x1=-1 x2=-6 
Ответ: х=-1