$23)z=2+3i$
Phần thực bằng $2$, phần ảo bằng $3$
$\Rightarrow D\\ 24)(2x+1)+(3y-2)i=(x+2)+(y+4)i\\ \Leftrightarrow \left\{\begin{array}{l} 2x+1=x+2\\3y-2=y+4\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=1\\y=3\end{array} \right.\\ \Rightarrow C\\ 25)\dfrac{3+2i}{2+3i}=\dfrac{(3+2i)(2-3i)}{(2+3i)(2-3i)}=\dfrac{12−5i}{13}=\dfrac{12}{13}-\dfrac{5}{13}i\\ \Rightarrow A\\ 26)\overrightarrow{AB}=(x_B-x_A;y_B-y_A;z_B-z_A)=(5;4;-4)\\ \Rightarrow C\\ 27)G=\left(\dfrac{x_A+x_B+x_C}{3};\dfrac{y_A+y_B+y_C}{3};\dfrac{z_A+z_B+z_C}{3}\right)=(1;0;2)\\ \Rightarrow C$
