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.
4)
4x-x ≥12+3
3x≥15
x≥5
4x+x≥18-3
5x≥15
x≥3
Ответ:x≥5
2^x+2/2^x -3≤0
2^2x-3*2^x+2/2^x≤0
2^x>0 при любом х⇒
2^2x-3*2^x+2≤0
2^x=a
a²-3a+2≤0
a1+a2=3 U a1*a2=2⇒a1=1 U a2=2
1≤a≤2⇒1≤2^x≤3⇒0≤x≤log(2)3
x∈[0;log(2)3]
A) 20-5x≥0, 5x≤20, x≤4, x∈(-∞; 4],
Ответ: D(y)=(-∞; 4]
b)10-2x≥0, 2x≤10, x≤5, x∈(-∞;5]
x+1≥0, x≥-1, x∈[-1; +∞)
(-∞;5]∩[-1; +∞)=[-1; 5]
Ответ: D(y)=[-1; 5]