3y-x=10
x=10+y
y^2-y(10+3y)=12
y^2-10y-3y^2=12
-2y^2-10y=12
2y^2+10y-12=0
y^2+5y-6=0
y1=-6;y2=1
подставим в формулу x1=10+(-6)=4;x2=10+1=11
надеюсь вы хоть что-нибудь поняли.давно не решала уравнения
![\alpha (x)=lncos5x-lncos2x\\\beta (x)=x](https://tex.z-dn.net/?f=%5Calpha+%28x%29%3Dlncos5x-lncos2x%5C%5C%5Cbeta+%28x%29%3Dx)
Найдем такое n, что
![\lim\limits_{x\to 0}=\frac{\alpha(x)}{\beta^n(x)} =\lim\limits_{x\to 0}\frac{lncos5x-lncos2x}{x^n} =A\neq 0\neq \infty](https://tex.z-dn.net/?f=%5Clim%5Climits_%7Bx%5Cto+0%7D%3D%5Cfrac%7B%5Calpha%28x%29%7D%7B%5Cbeta%5En%28x%29%7D+%3D%5Clim%5Climits_%7Bx%5Cto+0%7D%5Cfrac%7Blncos5x-lncos2x%7D%7Bx%5En%7D+%3DA%5Cneq+0%5Cneq+%5Cinfty)
Поехали:
![\lim\limits_{x\to 0}\frac{lncos5x-lncos2x}{x^n} =\lim\limits_{x\to 0}\frac{ln\frac{cos5x}{cos2x}}{x^n} =\lim\limits_{x\to 0}\frac{ln(1+\frac{cos5x}{cos2x}-1)}{x^n}=(*)](https://tex.z-dn.net/?f=%5Clim%5Climits_%7Bx%5Cto+0%7D%5Cfrac%7Blncos5x-lncos2x%7D%7Bx%5En%7D+%3D%5Clim%5Climits_%7Bx%5Cto+0%7D%5Cfrac%7Bln%5Cfrac%7Bcos5x%7D%7Bcos2x%7D%7D%7Bx%5En%7D+%3D%5Clim%5Climits_%7Bx%5Cto+0%7D%5Cfrac%7Bln%281%2B%5Cfrac%7Bcos5x%7D%7Bcos2x%7D-1%29%7D%7Bx%5En%7D%3D%28%2A%29)
ln(1+α)∼α, при α->0, поэтому
![(*)=\lim\limits_{x\to 0}\frac{\frac{cos5x}{cos2x}-1}{x^n}=\lim\limits_{x\to 0}\frac{cos5x-cos2x}{x^n}=-2\lim\limits_{x\to 0}\frac{sin\frac{7x}{2}sin\frac{3x}{2}} {x^n}=(*)](https://tex.z-dn.net/?f=%28%2A%29%3D%5Clim%5Climits_%7Bx%5Cto+0%7D%5Cfrac%7B%5Cfrac%7Bcos5x%7D%7Bcos2x%7D-1%7D%7Bx%5En%7D%3D%5Clim%5Climits_%7Bx%5Cto+0%7D%5Cfrac%7Bcos5x-cos2x%7D%7Bx%5En%7D%3D-2%5Clim%5Climits_%7Bx%5Cto+0%7D%5Cfrac%7Bsin%5Cfrac%7B7x%7D%7B2%7Dsin%5Cfrac%7B3x%7D%7B2%7D%7D+%7Bx%5En%7D%3D%28%2A%29)
sinα∼α, при α->0:
![(*)=-\frac{21}{2} \lim\limits_{x\to 0}\frac{x^2} {x^n}](https://tex.z-dn.net/?f=%28%2A%29%3D-%5Cfrac%7B21%7D%7B2%7D+%5Clim%5Climits_%7Bx%5Cto+0%7D%5Cfrac%7Bx%5E2%7D+%7Bx%5En%7D)
Из последнего равенства очевидно, что n=2. Итак, α(x) - бесконечно малая порядка 2 относительно β(x)