$\lim\limits_{x\to \pm\infty}y=1$
$\to$ TCN: $(a): y=1$ hay $y-1=0$
$\lim\limits_{x\to 3^+}y=+\infty$
$\to$ TCĐ: $(b): x=3$ hay $x-3=0$
Gọi $M\left(t; \dfrac{t+2}{t-3}\right)$ là điểm cần tìm
$d(M;a)=\dfrac{\left| \dfrac{t+2}{t-3} -1\right| }{1}=\left| \dfrac{t+2-t+3}{t-3}\right|=\left|\dfrac{5}{t-3}\right|$
$d(M;b)=\dfrac{|t-3|}{1}=|t-3|$
Ta có: $d(M;a)=d(M;b)$
$\to \dfrac{5}{|t-3|}=|t-3|$
$\to (t-3)^2=5$
$\to t=3\pm\sqrt5$ (TM $t\ne 3$)
Vậy $\left[ \begin{array}{l}M(3+\sqrt5; 1+\sqrt5) \\M(3-\sqrt5; 1-\sqrt5)\end{array} \right.$
