Goi M co toa do la $M(x;y),x>0$
Ta co: $\overrightarrow{AM}=(x-1;y)\Rightarrow AM=\sqrt{(x-1)^2+y^2}=\sqrt{x^2+y^2-2x+1}$
$\overrightarrow{BM}=(x;y-1)\Rightarrow BM=\sqrt{x^2+y^2-2y+1}$
theo bai ra ta co:$\left\{ \begin{array}{l} AM=5\\ BM=5 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} \sqrt{x^2+y^2-2x+1}=5\\ \sqrt{y^2+x^2-2y+1}=5 \end{array} \right.$ day la hpt dx loai 2
giai he tren duok nghiem $\left\{ \begin{array}{l} x=4\\ y=4 \end{array} \right.$
Vay $M(4;4)$