Đáp án:

$\begin{array}{l}\rm Bài\,\,1:\\a)\quad z = \cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\\b)\quad z^{2018} = \cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3}\\\rm Bài\,\,2:\\+)\quad \left[\begin{array}{l}m =0\\m =3 \end{array}\right.\Rightarrow r(A) = 2\\+)\quad \begin{cases}m \ne 0\\m \ne 3\end{cases}\Rightarrow r(A) = 3\\\rm Bài\,\,3:\\(x;y;z;t)=(0;1;1;1)\end{array}$

Giải thích các bước giải:

$\begin{array}{l}\rm Bài\,\,1:\\ \quad z = \dfrac{2i}{\sqrt3 + i}\\ \to z = \dfrac{2i(\sqrt3 - i)}{(\sqrt3 + i)(\sqrt3 - i)}\\ \to z = \dfrac12 + \dfrac{\sqrt3}{2}i\\ a)\quad \text{Ta có:}\\ \quad r = \sqrt{\left(\dfrac12\right)^2 + \left(\dfrac{\sqrt3}{2}\right)^2} = 1\\ \quad \varphi:\begin{cases}\cos\varphi = \dfrac12\\\sin\varphi = \dfrac{\sqrt3}{2}\end{cases}\to \varphi = \dfrac{\pi}{3} + k2\pi\quad (k\in\Bbb Z)\\ \text{Ta được:}\\ \quad z = \cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\\ b)\quad \text{Áp dụng công thức $Moivre$ ta được:}\\ \quad z^{2018} = \cos\dfrac{2018\pi}{3} + i\sin\dfrac{2018\pi}{3}\\ \to z^{2018} = \cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3} =-\dfrac12 + \dfrac{\sqrt3}{2}i\\ \rm Bài\,\,2:\\ \qquad A = \left(\matrix{4& m^2 -3m + 7 &6&5\cr 2&5&4&3\cr 3&6&5&4\cr 1&4&3&2} \right)\\ \xrightarrow{\quad r_4 \leftrightarrow r_1\quad }\left(\matrix{2&5&4&3\cr 3&6&5&4\cr 1&4&3&2\cr 4& m^2 -3m + 7 &6&5} \right)\\ \xrightarrow{\quad c_4 \leftrightarrow c_2\quad }\left(\matrix{2&3&4&5\cr 3&4&5&6\cr 1&2&3&4\cr 4& 5&6&m^2 -3m + 7 } \right)\\ \xrightarrow{\quad r_3 \leftrightarrow r_1\quad }\left(\matrix{ 1&2&3&4\cr 3&4&5&6\cr2&3&4&5\cr 4& 5&6&m^2 -3m + 7 } \right)\\ \xrightarrow{\begin{array}{l}r_2 -3r_1 \to r_2\\r_3 -r_1 \to r_3\\r_4 -4r_1 \to r_4\end{array}}\left(\matrix{ 1&2&3&4\cr 0&-2&-4&-6\cr0&-1&-2&-3\cr 0& -3&-6&m^2 -3m -9 } \right)\\ \xrightarrow{\begin{array}{l}r_3 -2r_2 \to r_3\\r_4 -\tfrac32r_2 \to r_4\end{array}}\left(\matrix{ 1&2&3&4\cr 0&-2&-4&-6\cr0&0&0&0\cr 0& 0&0&m^2 -3m } \right)\\ \xrightarrow{\qquad \qquad} \left(\matrix{ 1&2&3&4\cr 0&-2&-4&-6\cr 0& 0&0&m^2 -3m } \right)\\ +)\quad m^2 -3m =0\\ \Leftrightarrow \left[\begin{array}{l}m =0\\m = 3\end{array}\right.\\ \Rightarrow r(A) = 2\\ +)\quad \begin{cases}m \ne 0\\m \ne 3\end{cases}\\ \Rightarrow r(A) = 3\\ \rm Bài\,\,3:\\ \quad \begin{cases}2x + 2y - z + t =2\\x-y-z-t= -3\\x+y-z-2t=-2\\3x+y+z-5t =-3 \end{cases}\\ \qquad A = \left(\matrix{2&2&-1&1\cr 1&-1&-1&-1\cr 1&1&-1&-2\cr 3&1&1&-5} \right)\\ \Rightarrow \overline{A} = \left(\matrix{2&2&-1&1&2\cr 1&-1&-1&-1&-3\cr 1&1&-1&-2&-2\cr 3&1&1&-5&-3} \right)\\ \xrightarrow{\begin{array}{l} r_2 - \tfrac12r_1 \to r_2\\r_3 -\tfrac12r_1 \to r_3\\r_4 -\tfrac32r_1 \to r_4\end{array}}\left(\matrix{2&2&-1&1&2\cr 0&-2&-\dfrac12&-\dfrac32&-4\cr 0&0&-\dfrac12&-\dfrac52&-3\cr 0&-2&\dfrac52&-\dfrac{13}{2}&-6} \right)\\ \xrightarrow{r_4 -r_2 \to r_4}\left(\matrix{2&2&-1&1&2\cr 0&-2&-\dfrac12&-\dfrac32&-4\cr 0&0&-\dfrac12&-\dfrac52&-3\cr 0&0&3&-5&-2} \right)\\ \xrightarrow{r_4 + 6r_3 \to r_4}\left(\matrix{2&2&-1&1&2\cr 0&-2&-\dfrac12&-\dfrac32&-4\cr 0&0&-\dfrac12&-\dfrac52&-3\cr 0&0&0&-20&-20} \right)\\ \Rightarrow r(A) = r(\overline{A}) =4\\ \Rightarrow \text{Hệ phương trình có nghiệm duy nhất}\\ \text{Ta được:}\\ \quad \begin{cases}\,\,2x + 2y -z +t =2\\-2y - \dfrac12z -\dfrac32t = -4\\\qquad \,\,-\dfrac12z - \dfrac52t = -3\\\qquad \qquad \quad-20t = -20 \end{cases}\Leftrightarrow \begin{cases}x =0\\y=1\\z=1\\t=1\end{cases}\\ \text{Vậy hệ phương trình có nghiệm:}\,\,(x;y;z;t)=(0;1;1;1) \end{array}$