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Решите задачу:

$\displaystyle\\1)e^{i\pi}=-1\\\\e^{ix}=\cos(x)+i\sin(x)\\\\\frac{1}{3}e^{i\pi}*12e^{i*(-\frac{\pi}{4})}=-\frac{1}{3}*12(\cos(-\frac{\pi}{4})+i\sin(-\frac{\pi}{4}))=-4(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=\\\\\\=-2\sqrt{2}+2\sqrt{2}i\\\\\\2)\Delta=\left[\begin{array}{ccc}4&1&{-3}\\8&3&-6\\1&1&-1\end{array}\right]=4(-3+6)-1(-8+6)-3(8-3)=12+2-15=-1\\\\\\\Delta_1=\left[\begin{array}{ccc}-1&1&{-3}\\-1&3&-6\\-1&1&-1\end{array}\right]=-1(-3+6)-1(1-6)-3(-1+3)=-3+5-6=-4\\\\\\$

$\displaystyle\\\Delta_2=\left[\begin{array}{ccc}4&-1&{-3}\\8&-1&-6\\1&-1&-1\end{array}\right]=4(1-6)+1(-8+6)-3(-8+1)=-20-2+21=-1\\\\\\\Delta_3=\left[\begin{array}{ccc}4&1&{-1}\\8&3&-1\\1&1&-1\end{array}\right]=4(-3+1)-1(-8+1)-1(8-3)=-8+7-5=-6\\\\\\x_1=\frac{\Delta_1}{\Delta}=\frac{4}{1}=4\\\\\\x_2=\frac{\Delta_2}{\Delta}=\frac{1}{1}=1\\\\\\x_3=\frac{\Delta_3}{\Delta}=\frac{6}{1}=6$

$\displaystyle\\3)(x+1)dy-ydx=0\\\\\\(x+1)\frac{dy}{dx}-y=0\\\\\\\frac{\frac{dy}{dx} }{y}=\frac{1}{x+1}\\\\\\ \int\frac{\frac{dy}{dx} }{y}dx=\int\frac{1}{x+1}dx\\\\\\ \ln(y)=\ln(x+1)+C\\\\y=C(x+1)$