=1--12+1--3*(1.05-1.8)=1/12+1/3*(-0.75)=1--12+1--3*(-3--4)=1/12-1/4=-1--6
![\sqrt{5x^2+x} \geq 3x-1](https://tex.z-dn.net/?f=+%5Csqrt%7B5x%5E2%2Bx%7D++%5Cgeq+3x-1)
ОДЗ:
![5x^2+x \geq 0 \\ x(5x+1) \geq 0](https://tex.z-dn.net/?f=5x%5E2%2Bx+%5Cgeq+0+%5C%5C+x%285x%2B1%29+%5Cgeq+0)
a>0 ⇒ x∈(-∞;-1/5]U[0;+∞)
![5x^2+x \geq 9x^2-6x+1 \\ 4x^2-7x+1 \leq 0 \\ \\ 4x^2-7x+1=0 \\ D=49-16=33 \\ x_1= \dfrac{7- \sqrt{33} }{8} \\ x_2= \dfrac{7+ \sqrt{33} }{8}](https://tex.z-dn.net/?f=5x%5E2%2Bx+%5Cgeq+9x%5E2-6x%2B1+%5C%5C+4x%5E2-7x%2B1+%5Cleq+0+%5C%5C++%5C%5C+4x%5E2-7x%2B1%3D0+%5C%5C+D%3D49-16%3D33+%5C%5C+x_1%3D+%5Cdfrac%7B7-+%5Csqrt%7B33%7D+%7D%7B8%7D++%5C%5C+x_2%3D+%5Cdfrac%7B7%2B+%5Csqrt%7B33%7D+%7D%7B8%7D+)
Т.к.
![\sqrt{5x^2+x} = 3x-1](https://tex.z-dn.net/?f=%5Csqrt%7B5x%5E2%2Bx%7D+%3D+3x-1)
и
![\sqrt{5x^2+x} \geq 0](https://tex.z-dn.net/?f=+%5Csqrt%7B5x%5E2%2Bx%7D+%5Cgeq+0+)
то
![3x-1 \geq 0 \\ x \geq \dfrac{1}{3}](https://tex.z-dn.net/?f=3x-1+%5Cgeq+0+%5C%5C+x+%5Cgeq++%5Cdfrac%7B1%7D%7B3%7D+)
значит x=7-√33/8 - не точка смены знака в решении неравенства
С учетом ОДЗ
x∈(-∞;-1/5]U[0;7+√33/8]
Ответ: x∈(-∞;-1/5]U[0;7+√33/8]
-0.1x(2x2+6)(5-4x2)
-0.1x×(4+6)×(5-8)
-0.1x×10×(-3)
3x
Вроде так.
Все на картинке написано, ответ 2