Task/26898637
-------------------
а)
(2x -1) / (x+2) >0 ⇔2(x+2)(x-1/2) >0 .
x ∈( -∞ ; -2) ∪ (0,5 ; +∞) .
-------
б)
(14+5x) / (20 -x) < 0 ⇔5(x+14/5) / (x-20) >0 .
x ∈( -∞ ; -2,8) ∪ (20 ; +∞) .
-------
в)
(2x -15) /(17x+6) ≤ 0 ⇔(2/17)*(x -7,5) /(x +6/17) ≤0 ⇔(x -7,5) /(x +6/17) ≤ 0.
x∈ (-6/17 ; 7,5] .
-------
г)
- (1+6x)/ 4x ≥0 ⇔(6/4)*(x +1/6) / x ≤ 0⇔(x +1/6) / x ≤0 .
x ∈ [-1/6 ; 0) .
Ответ Б. у=2х
Но это тоже самое как ответ В. у=х2
Будет одинаковый график
Преобразуем исходное выражение:
![f(x)=1+tg^{2}4x=1+ \frac{sin^{2}4x}{cos^{2}4x} =1+ \frac{1-cos^{2}4x}{cos^{2}4x} =1+ \frac{1}{cos^{2}4x} -1=\frac{1}{cos^{2}4x}](https://tex.z-dn.net/?f=f%28x%29%3D1%2Btg%5E%7B2%7D4x%3D1%2B+%5Cfrac%7Bsin%5E%7B2%7D4x%7D%7Bcos%5E%7B2%7D4x%7D+%3D1%2B+%5Cfrac%7B1-cos%5E%7B2%7D4x%7D%7Bcos%5E%7B2%7D4x%7D+%3D1%2B+%5Cfrac%7B1%7D%7Bcos%5E%7B2%7D4x%7D+-1%3D%5Cfrac%7B1%7D%7Bcos%5E%7B2%7D4x%7D)
А теперь ищем первообразную:
![\int\limits {(1+tg^{2}4x)} \, dx =\int\limits {\frac{1}{cos^{2}4x}} \, dx = \frac{1}{4} \int\limits {\frac{1}{cos^{2}4x}} \, d(4x)= \frac{1}{4} tg(4x)+C](https://tex.z-dn.net/?f=+%5Cint%5Climits+%7B%281%2Btg%5E%7B2%7D4x%29%7D+%5C%2C+dx+%3D%5Cint%5Climits+%7B%5Cfrac%7B1%7D%7Bcos%5E%7B2%7D4x%7D%7D+%5C%2C+dx+%3D+%5Cfrac%7B1%7D%7B4%7D+%5Cint%5Climits+%7B%5Cfrac%7B1%7D%7Bcos%5E%7B2%7D4x%7D%7D+%5C%2C+d%284x%29%3D+%5Cfrac%7B1%7D%7B4%7D+tg%284x%29%2BC)
![\frac{1}{4}d(4x)=dx](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7Dd%284x%29%3Ddx)