$1.\\y=\dfrac{1}{3}x^3+x^2+3x-4\\\text{TXĐ: D=R}\\y'=x^2+2x+3=(x+1)^2+2>0∀∈R\\⇒\text{ Hàm số đồng biến trên R}\\⇒Miny_{[0;3]}=y(0)=\dfrac{1}{3}.0+0+3.0-4=-4\\Maxy_{[0;3]}=y(3)=\dfrac{1}{3}.3^3+3^2+3.3-4=23\\2.\\y=\dfrac{x^2-2x+3}{x+1}\\\text{TXĐ: D=R\ {-1}}\\ y'=\dfrac{(x^2-2x+3)'.(x+1)-(x^2-2x+3).(x+1)'}{(x+1)^2}\\=\dfrac{(2x-2).(x+1)-(x^2-2x+3).1}{(x+1)^2}\\=\dfrac{2x^2+2x-2x-2-x^2+2x-3}{(x+1)^2}\\=\dfrac{x^2+2x-5}{(x+1)^2}\\y'=0⇔x^2+2x-5=0⇔$\(\left[ \begin{array}{l}x=-1+\sqrt{6}\text{ thỏa mãn})\\x=-1-\sqrt{6}(\text{ loại vì: $-1-\sqrt{6}∉[0;2]$})\end{array} \right.\)
$y(0)=\dfrac{0-2.0+3}{0+1}=3\\y(2)=\dfrac{2^2-2.2+3}{2+1}=1\\y(-1+\sqrt{6})=\dfrac{(-1+\sqrt{6})^2-2.(-1+\sqrt{6})+3}{-1+\sqrt{6}+1}=-4+2\sqrt{6}$
$⇒Miny_{[0;2]}=-4+2\sqrt{6}\\Maxy_{[0;2]}=3$
