<span>1) (0,1-х)(х+0,1)=-(x-0.1)(x+0.1)=-(x²-0.01)=0.01-x²
2)</span><span> (1,2-а)(1,2+а)</span>=1.44-a²
3) <span> (1/3-m)(1/2m-3)</span>=-1/2m²+1/6m+3m-1=-1/2m²+19/6m-1=-1/6(3m²-19m+6)
0,6x-5,4=4,6-0,4x
0,6x+0,4x=4,6+5,4
x=10
![6cos^2x-7cosx-5=0; \\ (1+2cosx)(3cosx-5)=0; \\ 1)1+2cosx=0; \ 2cosx=-1 \\ cosx=- \frac{1}{2} ;\ x_1= \frac{2\pi}{3}+2\pi n ; x_2= \frac{4\pi}{3}+2\pi n; \\](https://tex.z-dn.net/?f=6cos%5E2x-7cosx-5%3D0%3B+%5C%5C+%281%2B2cosx%29%283cosx-5%29%3D0%3B+%5C%5C+1%291%2B2cosx%3D0%3B+%5C+2cosx%3D-1+%5C%5C+cosx%3D-+%5Cfrac%7B1%7D%7B2%7D+%3B%5C+x_1%3D+%5Cfrac%7B2%5Cpi%7D%7B3%7D%2B2%5Cpi+n+%3B+x_2%3D+%5Cfrac%7B4%5Cpi%7D%7B3%7D%2B2%5Cpi+n%3B+%5C%5C+)
2) 3cosx-5=0; 3cosx=5; cosx=5/3; x>1 => нет корней.
2п/3 ∉ [п,2п]
4п/3 ∈ [п,2п]
Ответ:
![x_1= \frac{4\pi}{3}](https://tex.z-dn.net/?f=x_1%3D+%5Cfrac%7B4%5Cpi%7D%7B3%7D+)