ta có $y'=\dfrac{(x^2-5)'.(x+3)-(x+3)'.(x^2-5)}{(x+3)^2}$
<=>$y'=\dfrac{2x(x+3)-(x^2-5)}{(x+3)^2}$
<=>$y'=\dfrac{2x^2+6x-x^2+5}{(x+3)^2}$
<=>$y'=\dfrac{x^2+6x+5}{(x+3)^2}=0$
=>\(\left[ \begin{array}{l}x=-1\\x=-5\end{array} \right.\)
ta có $y(-1)=-2$(L)
$y(-5)=-10$(L)
$y(0)=\dfrac{-5}{3}$
$y(2)=\dfrac{-1}{5}$
ta so sánh thì
$min_{[0;2]} y=-5/3$