Здесь будут использоваться замечательные пределы
![\displaystyle \lim_{x \to 0}\dfrac{a^x-1}{x}=\ln a;~~\lim_{x \to 0}\frac{{\rm tg}\, x}{x}=\lim_{x \to 0}\frac{{\rm arctg}\, x}{x}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%200%7D%5Cdfrac%7Ba%5Ex-1%7D%7Bx%7D%3D%5Cln%20a%3B~~%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7B%7B%5Crm%20tg%7D%5C%2C%20x%7D%7Bx%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7B%7B%5Crm%20arctg%7D%5C%2C%20x%7D%7Bx%7D%3D1)
![\displaystyle \lim_{x \to 0}\frac{10^{2x}-7^{-x}}{2{\rm tg}\, x-{\rm arctg}\, x}=\lim_{x \to 0}\frac{100^x-\dfrac{1}{7^x}}{2{\rm tg}\, x-{\rm arctg}\, x}=\lim_{x \to 0}\frac{\dfrac{1}{7^x}\left(700^x-1\right)}{2{\rm tg}\, x-{\rm arctg}\, x}=\\ \\ \\ =\lim_{x \to 0}\frac{1\cdot x \cdot (700^x-1)}{x\left(2{\rm tg}\, x-{\rm arctg}\, x\right)}=\lim_{x \to 0}\frac{x\ln700}{2{\rm tg}\, x-{\rm arctg}\, x}=\\ \\ \\ =\ln 700\lim_{x \to 0}\frac{1}{\dfrac{2{\rm tg}\, x}{x}-\dfrac{{\rm arctg}\, x}{x}}=\ln 700\cdot\frac{1}{2\cdot 1-1}=\ln700](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7B10%5E%7B2x%7D-7%5E%7B-x%7D%7D%7B2%7B%5Crm%20tg%7D%5C%2C%20x-%7B%5Crm%20arctg%7D%5C%2C%20x%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7B100%5Ex-%5Cdfrac%7B1%7D%7B7%5Ex%7D%7D%7B2%7B%5Crm%20tg%7D%5C%2C%20x-%7B%5Crm%20arctg%7D%5C%2C%20x%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7B%5Cdfrac%7B1%7D%7B7%5Ex%7D%5Cleft%28700%5Ex-1%5Cright%29%7D%7B2%7B%5Crm%20tg%7D%5C%2C%20x-%7B%5Crm%20arctg%7D%5C%2C%20x%7D%3D%5C%5C%20%5C%5C%20%5C%5C%20%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7B1%5Ccdot%20x%20%5Ccdot%20%28700%5Ex-1%29%7D%7Bx%5Cleft%282%7B%5Crm%20tg%7D%5C%2C%20x-%7B%5Crm%20arctg%7D%5C%2C%20x%5Cright%29%7D%3D%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7Bx%5Cln700%7D%7B2%7B%5Crm%20tg%7D%5C%2C%20x-%7B%5Crm%20arctg%7D%5C%2C%20x%7D%3D%5C%5C%20%5C%5C%20%5C%5C%20%3D%5Cln%20700%5Clim_%7Bx%20%5Cto%200%7D%5Cfrac%7B1%7D%7B%5Cdfrac%7B2%7B%5Crm%20tg%7D%5C%2C%20x%7D%7Bx%7D-%5Cdfrac%7B%7B%5Crm%20arctg%7D%5C%2C%20x%7D%7Bx%7D%7D%3D%5Cln%20700%5Ccdot%5Cfrac%7B1%7D%7B2%5Ccdot%201-1%7D%3D%5Cln700)
Ответ: ln700.
Область определения равна
(х>_4)..........
(4;+бесконечности)
Решаем уравнением
Х- скорость первого
Х+25-скорость второго
(Х+Х+25)*2=350
4Х+50=350
4Х=300
Х=300/4
Х=75
Х+25=100
<span>х^2 + х - 2корень из х^2 +х +4 = 4
</span>x^2+x-2x+2=4
x^2+x-2=0
по т.Виета
x1+x2=1
x1*x2=-2
x1=1
x2=-2
По теореме Виета
х₁+х₂=-b/2
х₁·х₂=c/2
t₁+t₂=3x₁+3x₂=3(x₁+x₂)=-3b/2
t₁·t₂=3x₁·3x₂=9x₁x₂=9c/2
Уравнение
t^2+(3b/2)t+(9c/2)=0
или
2t^2+3bt+9c=0
корни этого уравнения
t₁=3x₁; t₂=3x₂
О т в е т.
2x²+3bx+9x=0