Đặt z=a+bi; $\left( a;b\in \mathbb{R} \right)$

Ta có $\left( 2+3i \right)\left( z+2i-1 \right)=\left( 2i+1 \right)z$

$\begin{align} & \Leftrightarrow \left( 2+3i \right)\left[ a-1+\left( b+2 \right)i \right]=\left( 2i+1 \right)\left( a+bi \right) \\ & \Leftrightarrow \left( 2a-3b-7 \right)+\left( 3a+2b+1 \right)i=a-2b+\left( 2a+b \right)i \\ \end{align}$

$\Leftrightarrow \left\{ \begin{align} & 2a-3b-7=a-2b \\ & 3a+2b+1=2a+b \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align} & a-b=7 \\ & a+b=-1 \\ \end{align} \right.$ $\Leftrightarrow \left\{ \begin{align} & a=3 \\ & b=-4 \\ \end{align} \right.$

Vậy z=3-4i

Đáp án đúng là B