а)1/5x+4/5=15/5 перемножили скобку после равно привели к общему знаменателю
1/5x=15/5 - 4/5 записать дробью
1/5x=11/5
x = 11 * 5
----------
5
x=11
![2\arcsin(x)=\arccos(3x)](https://tex.z-dn.net/?f=2%5Carcsin%28x%29%3D%5Carccos%283x%29)
Для начала разберемся какому промежутку будет принадлежать корень уравнения.
![y=\arcsin(x)=>x\in\bigg[-1;1\bigg]\\y=\arccos(3x)\\-1\leq3x\leq 1=>x\in\bigg[-\frac{1}{3};\frac{1}{3}\bigg]](https://tex.z-dn.net/?f=y%3D%5Carcsin%28x%29%3D%3Ex%5Cin%5Cbigg%5B-1%3B1%5Cbigg%5D%5C%5Cy%3D%5Carccos%283x%29%5C%5C-1%5Cleq3x%5Cleq%201%3D%3Ex%5Cin%5Cbigg%5B-%5Cfrac%7B1%7D%7B3%7D%3B%5Cfrac%7B1%7D%7B3%7D%5Cbigg%5D)
Объедения промежутки получаем, что
![x\in\bigg[-\frac{1}{3};\frac{1}{3}\bigg]](https://tex.z-dn.net/?f=x%5Cin%5Cbigg%5B-%5Cfrac%7B1%7D%7B3%7D%3B%5Cfrac%7B1%7D%7B3%7D%5Cbigg%5D)
Теперь приступим к решению
![2\arcsin(x)=\arccos(3x)\\2\arcsin(x)-\arccos(3x)=0\\\cos(2\arcsin(x)-\arccos(3x))=\cos(0)\\\cos(\arccos(3x)-2\arcsin(x))=1\\\cos(\arccos(3x))\cos(-2\arcsin(x))-\sin(\arccos(3x)\sin(-2\arcsin(x))=1\\\\\cos(\arccos(3x))=3x\\\cos(-2\arcsin(x))=1-2\sin^2(\arcsin(x))=1-2x^2\\\sin(\arccos(3x))=\sqrt{1-cos^2(\arccos(3x))}=\sqrt{1-9x^2}\\\sin(-2\arcsin(x))=2\sin(-\arcsin(x)\cos(\arcsin(x))=-2x\sqrt{1-x^2}\\\\3x(1-2x^2)+2x\sqrt{1-9x^2}\sqrt{1-x^2}=1\\\left(2x\sqrt{(1-9x^2)(1-x^2)}\right)^2=(1-3x+6x^3)^2](https://tex.z-dn.net/?f=2%5Carcsin%28x%29%3D%5Carccos%283x%29%5C%5C2%5Carcsin%28x%29-%5Carccos%283x%29%3D0%5C%5C%5Ccos%282%5Carcsin%28x%29-%5Carccos%283x%29%29%3D%5Ccos%280%29%5C%5C%5Ccos%28%5Carccos%283x%29-2%5Carcsin%28x%29%29%3D1%5C%5C%5Ccos%28%5Carccos%283x%29%29%5Ccos%28-2%5Carcsin%28x%29%29-%5Csin%28%5Carccos%283x%29%5Csin%28-2%5Carcsin%28x%29%29%3D1%5C%5C%5C%5C%5Ccos%28%5Carccos%283x%29%29%3D3x%5C%5C%5Ccos%28-2%5Carcsin%28x%29%29%3D1-2%5Csin%5E2%28%5Carcsin%28x%29%29%3D1-2x%5E2%5C%5C%5Csin%28%5Carccos%283x%29%29%3D%5Csqrt%7B1-cos%5E2%28%5Carccos%283x%29%29%7D%3D%5Csqrt%7B1-9x%5E2%7D%5C%5C%5Csin%28-2%5Carcsin%28x%29%29%3D2%5Csin%28-%5Carcsin%28x%29%5Ccos%28%5Carcsin%28x%29%29%3D-2x%5Csqrt%7B1-x%5E2%7D%5C%5C%5C%5C3x%281-2x%5E2%29%2B2x%5Csqrt%7B1-9x%5E2%7D%5Csqrt%7B1-x%5E2%7D%3D1%5C%5C%5Cleft%282x%5Csqrt%7B%281-9x%5E2%29%281-x%5E2%29%7D%5Cright%29%5E2%3D%281-3x%2B6x%5E3%29%5E2)
![4x^2(1-9x^2)(1-x^2)=(1-3x+6x^3)(1-3x+6x^3)\\36x^6-40x^4+4x^2=(1-3x+6x^3+9x-18x^4-6x^3-18x^4+36x^6)\\36x^6-40x^4+4x^2=36x^4-36x^4+12x^3+9x^2-6x+1\\4x^4+12x^3+5x^2-6x+1=0\\4x^4+(6x^3+6x^3)+(9x^2-4x^2)+(-3x-3x)+1=0\\2x^2(2x^2+3x-1)+3x(2x^2+3x-1)-(2x^2+3x-1)=0\\(2x^2+3x-1)(2x^2+3x-1)=0\\(2x^2+3x-1)^2=0\\2x^2+3x-1=0\\D=9-4*2*(-1)=17\\x_{1,2}=\frac{-3\pm\sqrt{17}}{4}](https://tex.z-dn.net/?f=4x%5E2%281-9x%5E2%29%281-x%5E2%29%3D%281-3x%2B6x%5E3%29%281-3x%2B6x%5E3%29%5C%5C36x%5E6-40x%5E4%2B4x%5E2%3D%281-3x%2B6x%5E3%2B9x-18x%5E4-6x%5E3-18x%5E4%2B36x%5E6%29%5C%5C36x%5E6-40x%5E4%2B4x%5E2%3D36x%5E4-36x%5E4%2B12x%5E3%2B9x%5E2-6x%2B1%5C%5C4x%5E4%2B12x%5E3%2B5x%5E2-6x%2B1%3D0%5C%5C4x%5E4%2B%286x%5E3%2B6x%5E3%29%2B%289x%5E2-4x%5E2%29%2B%28-3x-3x%29%2B1%3D0%5C%5C2x%5E2%282x%5E2%2B3x-1%29%2B3x%282x%5E2%2B3x-1%29-%282x%5E2%2B3x-1%29%3D0%5C%5C%282x%5E2%2B3x-1%29%282x%5E2%2B3x-1%29%3D0%5C%5C%282x%5E2%2B3x-1%29%5E2%3D0%5C%5C2x%5E2%2B3x-1%3D0%5C%5CD%3D9-4%2A2%2A%28-1%29%3D17%5C%5Cx_%7B1%2C2%7D%3D%5Cfrac%7B-3%5Cpm%5Csqrt%7B17%7D%7D%7B4%7D)
Осталось проверить принадлежать ли найденные корни найденному ранее промежутку. Получаем:
![x=\frac{-3+\sqrt{17}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-3%2B%5Csqrt%7B17%7D%7D%7B4%7D)
Готово, переносим все в одну сторону, кроме у, подставляем любые 2 числа под х , находим у и переносим все на график
Функция: y = 2х-9
A) y = 2 * (-2,5) - 9 = -5 -9 = -14
Б) 2х-9=5; 2х=5+9; 2х=14; x =7
B) да, проходит, проверка -39= 2*(-15)-9
Просмотрел ваш LaTeX код в общем виде, у Вас х стремится к -5.
![\displaystyle \lim_{x \to-5}\frac{\sqrt{5-4x}-5}{x^2-25}=\lim_{x \to-5}\frac{(\sqrt{5-4x}-5)(\sqrt{5-4x}+5)}{(x^2-25)(\sqrt{5-4x}+5)}=\\ \\ \\ =\lim_{x \to-5}\frac{5-4x-25}{(x-5)(x+5)(\sqrt{5-4x}+5)}=\lim_{x \to-5}\frac{-4(x+5)}{(x-5)(x+5)(\sqrt{5-4x}+5)}=\\ \\ \\ =-4\lim_{x \to-5}\frac{1}{(x-5)(\sqrt{5-4x}+5)}=-4\cdot\frac{1}{(-5-5)\cdot(\sqrt{5+20}+5)}=0.04](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto-5%7D%5Cfrac%7B%5Csqrt%7B5-4x%7D-5%7D%7Bx%5E2-25%7D%3D%5Clim_%7Bx%20%5Cto-5%7D%5Cfrac%7B%28%5Csqrt%7B5-4x%7D-5%29%28%5Csqrt%7B5-4x%7D%2B5%29%7D%7B%28x%5E2-25%29%28%5Csqrt%7B5-4x%7D%2B5%29%7D%3D%5C%5C%20%5C%5C%20%5C%5C%20%3D%5Clim_%7Bx%20%5Cto-5%7D%5Cfrac%7B5-4x-25%7D%7B%28x-5%29%28x%2B5%29%28%5Csqrt%7B5-4x%7D%2B5%29%7D%3D%5Clim_%7Bx%20%5Cto-5%7D%5Cfrac%7B-4%28x%2B5%29%7D%7B%28x-5%29%28x%2B5%29%28%5Csqrt%7B5-4x%7D%2B5%29%7D%3D%5C%5C%20%5C%5C%20%5C%5C%20%3D-4%5Clim_%7Bx%20%5Cto-5%7D%5Cfrac%7B1%7D%7B%28x-5%29%28%5Csqrt%7B5-4x%7D%2B5%29%7D%3D-4%5Ccdot%5Cfrac%7B1%7D%7B%28-5-5%29%5Ccdot%28%5Csqrt%7B5%2B20%7D%2B5%29%7D%3D0.04)