$y=\dfrac{x-2}{x+3}\\\to y'=\Bigg(\dfrac{x-2}{x-3}\Bigg)'=\dfrac{(x-2)'(x+3)-(x-2)(x+3)'}{(x+3)^2}=\dfrac{x+3-x+2}{(x+3)^2}=\dfrac{5}{(x+3)^2}\\d: -5x+y-2018=0 \to y=5x+2018 \to k=5\\\to \text{Hệ số góc của tiếp tuyến: }k=5\\\to \dfrac{5}{(x+3)^2}=5 \to 5(x+3)^2=5 \\\to (x+3)^2=1 \to \left[\begin{array}{l}x+3=1\\x+3=-1\end{array}\right. \to \left[\begin{array}{l}x=-2\\x=-4\end{array}\right.\\\to \left[\begin{array}{l}y=\dfrac{-2-2}{-2+3}=-4\\y=\dfrac{-4-2}{-4+3}=6\end{array}\right.\to \left[\begin{array}{l}M(-2; -4)\\M(-4; 6)\end{array}\right.\\\to C$
