<span>3 sin^2x-cosx+1=0</span>
по формуле sin^2(x) + cos^2(x) = 1
3 sin^2(x) = 3 - 3 cos^2(x)
3 - 3cos^2(x) - cos(x)+1=0
3cos^2(x) + cos(x) - 4 = 0
Дальше можно через дескриминант или по теореме Виетта
Я пойду первым способом
Заменим cos(x)=t
3t + t - 4=0
![t=\frac{1+-\sqrt{1+48}}{6}=\frac{1+-7}{6}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B1%2B-%5Csqrt%7B1%2B48%7D%7D%7B6%7D%3D%5Cfrac%7B1%2B-7%7D%7B6%7D)
=> t=4/3 и t=1
=> cos(x) = 4/3 - что не возможно и cos(x)=1
=> x=0
Вроде бы так)
Удачи)
Формула:
х-х1/у-у1=х2-х1/у2-у1
х+1/у-3=4/-12
4(у-3)=-12(х+1)
4у-12=-12х-12
4у+12х=0
1)
![\displaystyle \sqrt{6^4} = \sqrt{(6^2)^2}=6^2=36](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Csqrt%7B6%5E4%7D+%3D+%5Csqrt%7B%286%5E2%29%5E2%7D%3D6%5E2%3D36+)
2)
![\displaystyle \sqrt{(-5)^6}= \sqrt{(-1)^6*5^6}= \sqrt{1*(5^3)^2}=5^3=125](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Csqrt%7B%28-5%29%5E6%7D%3D+%5Csqrt%7B%28-1%29%5E6%2A5%5E6%7D%3D+%5Csqrt%7B1%2A%285%5E3%29%5E2%7D%3D5%5E3%3D125+)
Заметим что Четная степень отрицательного числа = положительному числу
3)
![\displaystyle \sqrt{(-8)^4}= \sqrt{(-1)^4*8^4}=8^2=64](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Csqrt%7B%28-8%29%5E4%7D%3D+%5Csqrt%7B%28-1%29%5E4%2A8%5E4%7D%3D8%5E2%3D64++)
4)
![\displaystyle \sqrt{0.1^6}= \sqrt{(0.1^3)^2}=0.1^3= 0.001](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Csqrt%7B0.1%5E6%7D%3D+%5Csqrt%7B%280.1%5E3%29%5E2%7D%3D0.1%5E3%3D+0.001)
5)
![\displaystyle \sqrt{(-1)^{4n}}= \sqrt{((-1)^2)^{2n}} =\sqrt{1^{2n}}= 1^{2n}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Csqrt%7B%28-1%29%5E%7B4n%7D%7D%3D+%5Csqrt%7B%28%28-1%29%5E2%29%5E%7B2n%7D%7D+%3D%5Csqrt%7B1%5E%7B2n%7D%7D%3D+1%5E%7B2n%7D%3D1+)
6)
![\displaystyle \sqrt{(-1)^{4n+6}}= \sqrt{((-1)^2)^{2n+3}}= \sqrt{1^{2n+3}}=1^{2n+3}=1](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Csqrt%7B%28-1%29%5E%7B4n%2B6%7D%7D%3D+%5Csqrt%7B%28%28-1%29%5E2%29%5E%7B2n%2B3%7D%7D%3D+%5Csqrt%7B1%5E%7B2n%2B3%7D%7D%3D1%5E%7B2n%2B3%7D%3D1+++)
3)
![\frac{9x+12}{x^3-64} - \frac{1}{x^2+4x+16} = \frac{1}{x-4}](https://tex.z-dn.net/?f=+%5Cfrac%7B9x%2B12%7D%7Bx%5E3-64%7D+-+%5Cfrac%7B1%7D%7Bx%5E2%2B4x%2B16%7D+%3D+%5Cfrac%7B1%7D%7Bx-4%7D+)
Область определения: x =/= 4.
Знаменатель x^3 - 64 = (x - 4)(x^2 + 4x + 16). Умножаем на него
9x + 12 - (x - 4) = x^2 + 4x + 16
x^2 + 4x + 16 - 8x - 16 = 0
x^2 - 4x = 0
x1 = 0; x2 = 4 - не подходит
Ответ: 0
4)
![\frac{x \sqrt{3}+ \sqrt{2}}{x \sqrt{3}- \sqrt{2}} + \frac{x \sqrt{3}- \sqrt{2}}{x \sqrt{3}+ \sqrt{2}} = \frac{10x}{3x^2-2}](https://tex.z-dn.net/?f=+%5Cfrac%7Bx+%5Csqrt%7B3%7D%2B+%5Csqrt%7B2%7D%7D%7Bx+%5Csqrt%7B3%7D-+%5Csqrt%7B2%7D%7D+%2B+%5Cfrac%7Bx+%5Csqrt%7B3%7D-+%5Csqrt%7B2%7D%7D%7Bx+%5Csqrt%7B3%7D%2B+%5Csqrt%7B2%7D%7D+%3D+%5Cfrac%7B10x%7D%7B3x%5E2-2%7D+)
Область определения: x =/= -√2/√3; x =/= √2/√3
Знаменатель 3x^2 - 2 = (x√3 - √2)(x√3 + √2). Умножаем на него.
(x√3 + √2)^2 + (x√3 - √2)^2 = 10x
3x^2 + 2x√6 + 2 + 3x^2 - 2x√6 + 2 = 10x
6x^2 + 4 = 10x
Делим всё на 2
3x^2 - 5x + 2 = 0
(x - 1)(3x - 2) = 0
x1 = 1; x2 = 2/3