Рассмотрим комплексное число:
![z=\sqrt{3}-i\sqrt{3}](https://tex.z-dn.net/?f=z%3D%5Csqrt%7B3%7D-i%5Csqrt%7B3%7D)
Найдем его модуль и аргумент:
![|z|=\sqrt{(\sqrt{3})^2+(\sqrt{3})^2}=\sqrt{6}](https://tex.z-dn.net/?f=%7Cz%7C%3D%5Csqrt%7B%28%5Csqrt%7B3%7D%29%5E2%2B%28%5Csqrt%7B3%7D%29%5E2%7D%3D%5Csqrt%7B6%7D)
![\arg z=\mathrm{arctg}\dfrac{-\sqrt{3}}{\sqrt{3}}=\mathrm{arctg}(-1)=-\dfrac{\pi }{4}](https://tex.z-dn.net/?f=%5Carg+z%3D%5Cmathrm%7Barctg%7D%5Cdfrac%7B-%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B3%7D%7D%3D%5Cmathrm%7Barctg%7D%28-1%29%3D-%5Cdfrac%7B%5Cpi+%7D%7B4%7D)
Запишем число в тригонометрической форме:
![z=\sqrt{6}\left(\cos\left(-\dfrac{\pi }{4} \right)+i\sin\left(-\dfrac{\pi }{4} \right)\right)](https://tex.z-dn.net/?f=z%3D%5Csqrt%7B6%7D%5Cleft%28%5Ccos%5Cleft%28-%5Cdfrac%7B%5Cpi+%7D%7B4%7D+%5Cright%29%2Bi%5Csin%5Cleft%28-%5Cdfrac%7B%5Cpi+%7D%7B4%7D+%5Cright%29%5Cright%29)
Найдем значения кубического корня:
![\sqrt[3]{\rho(\cos \phi+i\sin \phi)} =\left\{\sqrt[3]{\rho}\left(\cos\dfrac{\phi+2\pi k}{3} +i\sin\dfrac{\phi+2\pi k}{3} \right)|k=0;1;2\right\}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Crho%28%5Ccos+%5Cphi%2Bi%5Csin+%5Cphi%29%7D+%3D%5Cleft%5C%7B%5Csqrt%5B3%5D%7B%5Crho%7D%5Cleft%28%5Ccos%5Cdfrac%7B%5Cphi%2B2%5Cpi+k%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B%5Cphi%2B2%5Cpi+k%7D%7B3%7D+%5Cright%29%7Ck%3D0%3B1%3B2%5Cright%5C%7D)
![(\sqrt[3]{z})_1=\sqrt[3]{\sqrt{6}}\left(\cos\dfrac{-\frac{\pi }{4}}{3} +i\sin\dfrac{-\frac{\pi }{4}}{3} \right)=\sqrt[6]{6}\left(\cos\left(-\dfrac{\pi }{12}\right)+i\sin\left(-\dfrac{\pi }{12}\right)\right)](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bz%7D%29_1%3D%5Csqrt%5B3%5D%7B%5Csqrt%7B6%7D%7D%5Cleft%28%5Ccos%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%7D%7B3%7D+%5Cright%29%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cleft%28-%5Cdfrac%7B%5Cpi+%7D%7B12%7D%5Cright%29%2Bi%5Csin%5Cleft%28-%5Cdfrac%7B%5Cpi+%7D%7B12%7D%5Cright%29%5Cright%29)
![(\sqrt[3]{z})_2=\sqrt[3]{\sqrt{6}}\left(\cos\dfrac{-\frac{\pi }{4}+2\pi }{3} +i\sin\dfrac{-\frac{\pi }{4}+2\pi }{3} \right)=\\=\sqrt[6]{6}\left(\cos\dfrac{\frac{7\pi }{4} }{3} +i\sin\dfrac{\frac{7\pi }{4} }{3} \right)=\sqrt[6]{6}\left(\cos\dfrac{7\pi }{12}+i\sin\dfrac{7\pi }{12}\right)](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bz%7D%29_2%3D%5Csqrt%5B3%5D%7B%5Csqrt%7B6%7D%7D%5Cleft%28%5Ccos%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%2B2%5Cpi+%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%2B2%5Cpi+%7D%7B3%7D+%5Cright%29%3D%5C%5C%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B%5Cfrac%7B7%5Cpi+%7D%7B4%7D+%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B%5Cfrac%7B7%5Cpi+%7D%7B4%7D+%7D%7B3%7D+%5Cright%29%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B7%5Cpi+%7D%7B12%7D%2Bi%5Csin%5Cdfrac%7B7%5Cpi+%7D%7B12%7D%5Cright%29)
![(\sqrt[3]{z})_3=\sqrt[3]{\sqrt{6}}\left(\cos\dfrac{-\frac{\pi }{4}+4\pi }{3} +i\sin\dfrac{-\frac{\pi }{4}+4\pi }{3} \right)=\\=\sqrt[6]{6}\left(\cos\dfrac{\frac{15\pi }{4} }{3} +i\sin\dfrac{\frac{15\pi }{4} }{3} \right)=\sqrt[6]{6}\left(\cos\dfrac{15\pi }{12}+i\sin\dfrac{15\pi }{12}\right)=\\=\sqrt[6]{6}\left(\cos\dfrac{5\pi }{4}+i\sin\dfrac{5\pi }{3}\right)=\sqrt[6]{6}\left(\cos\left(-\dfrac{3\pi }{4}\right)+i\sin\left(-\dfrac{3\pi }{4}\right)\right)](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bz%7D%29_3%3D%5Csqrt%5B3%5D%7B%5Csqrt%7B6%7D%7D%5Cleft%28%5Ccos%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%2B4%5Cpi+%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%2B4%5Cpi+%7D%7B3%7D+%5Cright%29%3D%5C%5C%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B%5Cfrac%7B15%5Cpi+%7D%7B4%7D+%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B%5Cfrac%7B15%5Cpi+%7D%7B4%7D+%7D%7B3%7D+%5Cright%29%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B15%5Cpi+%7D%7B12%7D%2Bi%5Csin%5Cdfrac%7B15%5Cpi+%7D%7B12%7D%5Cright%29%3D%5C%5C%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B5%5Cpi+%7D%7B4%7D%2Bi%5Csin%5Cdfrac%7B5%5Cpi+%7D%7B3%7D%5Cright%29%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cleft%28-%5Cdfrac%7B3%5Cpi+%7D%7B4%7D%5Cright%29%2Bi%5Csin%5Cleft%28-%5Cdfrac%7B3%5Cpi+%7D%7B4%7D%5Cright%29%5Cright%29)
При изображении получившийся модуль числа
является длиной векторов, а получившиеся аргументы -п/12, 7п/12, -3п/4 - углами, на которые нужно повернуть ось х для ее совмещения с направлением векторов
А что именно? (3.54-3.59) = -0.05
При умножении степени складываем,а при делении степени отнимаем.
А) f`(x)=(x²)`=2x
<span>б) Вычислите значение производной в точке х=0; 1; -1; 2; -2; 3;-3
f`(0)=2·0=0
</span><span> f`(1)=2·1=2
f`(-1)=2·(-1)=-2
</span>f`(2)=2·2=4
<span>f`(-2)=2·(-2)=-4
</span><span>f`(3)=2·3=6
</span><span><span>f`(-3)=2·(-3)=-6
</span>в) При каком значении х производная равна: 0; 1; 3</span>
2x = 0 ⇒ x = 0
2x = 1 ⇒ x = 1/2
2x = 3 ⇒ x = 3/2=1,5
Полученая коробка это параллелепипед.
Объем параллелепипеда V=a*b*c
1 сторона a-2x
2 сторона b-2x
высота с
тогда^
V=(a-2x)*(b-2x)*c