Помогите по братски
Ответ:
Объяснение:
Найдем общие точки функций, то есть приравняем их и решим уравнение:
а можно по русски?
![\[(x;y) = \{ (1;10);( - 1; - 10)\} \]](https://tex.z-dn.net/?f=%5C%5B%28x%3By%29%20%3D%20%5C%7B%20%281%3B10%29%3B%28%20-%201%3B%20-%2010%29%5C%7D%20%5C%5D "\[(x;y) = \{ (1;10);( - 1; - 10)\} \]")
![\[\begin{array}{l}f(x) = 2x + \frac{8}{x}\\f(4x) = 2 \cdot 4x + \frac{8}{{4x}} = 8x + \frac{2}{x}\end{array}\]](https://tex.z-dn.net/?f=%5C%5B%5Cbegin%7Barray%7D%7Bl%7Df%28x%29%20%3D%202x%20%2B%20%5Cfrac%7B8%7D%7Bx%7D%5C%5Cf%284x%29%20%3D%202%20%5Ccdot%204x%20%2B%20%5Cfrac%7B8%7D%7B%7B4x%7D%7D%20%3D%208x%20%2B%20%5Cfrac%7B2%7D%7Bx%7D%5Cend%7Barray%7D%5C%5D "\[\begin{array}{l}f(x) = 2x + \frac{8}{x}\\f(4x) = 2 \cdot 4x + \frac{8}{{4x}} = 8x + \frac{2}{x}\end{array}\]")
![\[\begin{array}{l}2x + \frac{8}{x} = 8x + \frac{2}{x}|:2\\x + \frac{4}{x} = 4x + \frac{1}{x}\\x - \frac{1}{x} - 4x + \frac{4}{x} = 0\\ - 3x + \frac{3}{x} = 0|:3\\x - \frac{1}{x} = 0| \cdot x,x \ne 0\\{x^2} - 1 = 0\\(x + 1)(x - 1) = 0\\x = \pm 1\\f(1) = 10\\f( - 1) = - 10\\(x;y) = \{ (1;10);( - 1; - 10)\} \end{array}\]](https://tex.z-dn.net/?f=%5C%5B%5Cbegin%7Barray%7D%7Bl%7D2x%20%2B%20%5Cfrac%7B8%7D%7Bx%7D%20%3D%208x%20%2B%20%5Cfrac%7B2%7D%7Bx%7D%7C%3A2%5C%5Cx%20%2B%20%5Cfrac%7B4%7D%7Bx%7D%20%3D%204x%20%2B%20%5Cfrac%7B1%7D%7Bx%7D%5C%5Cx%20-%20%5Cfrac%7B1%7D%7Bx%7D%20-%204x%20%2B%20%5Cfrac%7B4%7D%7Bx%7D%20%3D%200%5C%5C%20-%203x%20%2B%20%5Cfrac%7B3%7D%7Bx%7D%20%3D%200%7C%3A3%5C%5Cx%20-%20%5Cfrac%7B1%7D%7Bx%7D%20%3D%200%7C%20%5Ccdot%20x%2Cx%20%5Cne%200%5C%5C%7Bx%5E2%7D%20-%201%20%3D%200%5C%5C%28x%20%2B%201%29%28x%20-%201%29%20%3D%200%5C%5Cx%20%3D%20%20%5Cpm%201%5C%5Cf%281%29%20%3D%2010%5C%5Cf%28%20-%201%29%20%3D%20%20-%2010%5C%5C%28x%3By%29%20%3D%20%5C%7B%20%281%3B10%29%3B%28%20-%201%3B%20-%2010%29%5C%7D%20%5Cend%7Barray%7D%5C%5D "\[\begin{array}{l}2x + \frac{8}{x} = 8x + \frac{2}{x}|:2\\x + \frac{4}{x} = 4x + \frac{1}{x}\\x - \frac{1}{x} - 4x + \frac{4}{x} = 0\\ - 3x + \frac{3}{x} = 0|:3\\x - \frac{1}{x} = 0| \cdot x,x \ne 0\\{x^2} - 1 = 0\\(x + 1)(x - 1) = 0\\x = \pm 1\\f(1) = 10\\f( - 1) = - 10\\(x;y) = \{ (1;10);( - 1; - 10)\} \end{array}\]")