![\frac{a^2b^2}{a^2+2ab+b^2}: \frac{ab}{a+b}= \frac{(ab)^2}{(a+b)^2}* \frac{a+b}{ab}= \frac{ab}{a+b}](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%5E2b%5E2%7D%7Ba%5E2%2B2ab%2Bb%5E2%7D%3A+%5Cfrac%7Bab%7D%7Ba%2Bb%7D%3D+%5Cfrac%7B%28ab%29%5E2%7D%7B%28a%2Bb%29%5E2%7D%2A+%5Cfrac%7Ba%2Bb%7D%7Bab%7D%3D+%5Cfrac%7Bab%7D%7Ba%2Bb%7D+)
<span>если а=4-√3, b=4+√3
</span>
![\frac{(4- \sqrt{3}) *(4+ \sqrt{3}) }{(4- \sqrt{3} )+(4+ \sqrt{3} )} = \frac{16-3}{8} = \frac{13}{8}=1,625](https://tex.z-dn.net/?f=%5Cfrac%7B%284-+%5Csqrt%7B3%7D%29+%2A%284%2B+%5Csqrt%7B3%7D%29+%7D%7B%284-+%5Csqrt%7B3%7D+%29%2B%284%2B+%5Csqrt%7B3%7D+%29%7D+%3D+%5Cfrac%7B16-3%7D%7B8%7D+%3D+%5Cfrac%7B13%7D%7B8%7D%3D1%2C625+)
x² + 4x + 3 = 0
x(x + 3) + x + 3 = 0
(x + 3)(x + 1) = 0
{x + 3 = 0 {x = 0 - 3 {x₁ = - 3
{x + 1 = 0 {x = 0 - 1 {x₂ = - 1
ИЛИ
D = b² - 4ac = 16 - 12 = 4
{x₁ = (- 4 + 2) : 2 = - 2 : 2 = - 1
{x₂ = (- 4 - 2) : 2 = - 6 : 2 = - 3
Пользуясь формулой, получаем: cos(pi/3 -3x)=cos(pi/3)*cos(3x)+sin(pi/3)*sin(3x)=1/2*cos(3x)+√3/2*sin(3x). Тогда первообразная будет равна: интеграл(1/2*cos(3x)+√3/2*sin(3x))dx=интеграл(1/2*cos(3x))dx + интеграл(√3/2*sin(3x))dx=1/2 интеграл(cos(3x))dx + √3/2 интеграл(sin(3x))dx=1/2*(sin(3x)/3) - √3/2*(cos(3x)/3) + C=(sin(3x) - √3cos(3x))/6 + C.