ОДЗ :
1) 8x ≥ 0
x ≥ 0
2) 3x² - 3 ≥ 0
x² - 1 ≥ 0
x ∈ (- ∞ ; - 1 ] ∪ [1 ; + ∞ )
Окончательно : x ∈ [1 , + ∞)
![3^{\sqrt{3x^{2}-3}}=3^{\sqrt{8x}}\\\sqrt{3x^{2} -3}=\sqrt{8x}\\\\(\sqrt{3x^{2}-3 })^{2}=(\sqrt{8x})^{2}\\\\3x^{2}-3=8x\\\\3x^{2}-8x-3=0\\\\D=(-8)^{2}-4*3*(-3)=64+36=100=10^{2}\\\\x_{1}=\frac{8+10}{6}=3\\\\x_{2}=\frac{8-10}{6}= -\frac{1}[3}](https://tex.z-dn.net/?f=3%5E%7B%5Csqrt%7B3x%5E%7B2%7D-3%7D%7D%3D3%5E%7B%5Csqrt%7B8x%7D%7D%5C%5C%5Csqrt%7B3x%5E%7B2%7D%20-3%7D%3D%5Csqrt%7B8x%7D%5C%5C%5C%5C%28%5Csqrt%7B3x%5E%7B2%7D-3%20%7D%29%5E%7B2%7D%3D%28%5Csqrt%7B8x%7D%29%5E%7B2%7D%5C%5C%5C%5C3x%5E%7B2%7D-3%3D8x%5C%5C%5C%5C3x%5E%7B2%7D-8x-3%3D0%5C%5C%5C%5CD%3D%28-8%29%5E%7B2%7D-4%2A3%2A%28-3%29%3D64%2B36%3D100%3D10%5E%7B2%7D%5C%5C%5C%5Cx_%7B1%7D%3D%5Cfrac%7B8%2B10%7D%7B6%7D%3D3%5C%5C%5C%5Cx_%7B2%7D%3D%5Cfrac%7B8-10%7D%7B6%7D%3D%20-%5Cfrac%7B1%7D%5B3%7D)
Ответ : 3
По формуле разности квадратов:
(10a-1)(10a+1)
А2+а4+а6=3*а4=33⇒а4=11
а2+а6=22
11*а2*а6=935⇒а2*а6=85
а2=5, а6=17, d=6, a1=-1
a1*d=-6
(30 * (0,03) + 70 * (0,04))/100 = 0,037
так должно быть
Х=1+у
(1+у)*у=12
у^2+y-12=0
D=1+4*12=49 ,больше 0 - 2 корня
у1= (-1+7)/2=3
у2=(-1-7)/2=-4
х1=1+3=4
х2=1-4=-3
Ответ: (4;3) и (-3;-4)