$A \in \Delta_1:2x+3y-5=0\\ \Rightarrow A\left(x_1;\dfrac{5-2x_1}{3}\right)\\ B \in \Delta_2:x+y-3=0\\ \Rightarrow B(x_2;3-x_2)$
$B$ là trung điểm $AM$
$\Rightarrow \left\{\begin{array}{l} x_1-2=2x_2\\ \dfrac{5-2x_1}{3}=2(3-x_2) \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x_1-2x_2=2\\-2x_1+6x_2=13\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x_1=19\\ x_2 =8,5\end{array} \right.\\ \Rightarrow A\left(19;-11\right);B(8,5;-5,5)$
Gọi đường thẳng cần tìm có dạng $y=ax+b(d)$
$d $ đi qua $A,M$
$\Rightarrow \left\{\begin{array}{l} 19a+b=-11\\-2a+b=0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} a=-\dfrac{11}{21}\\b=-\dfrac{22}{21}\end{array} \right.\\ \Rightarrow (d)y=-\dfrac{11}{21}x-\dfrac{22}{21}$
