`y=(x+3).\sqrt{-x^2-2x+3}`
TXĐ: `D=[-3;1]`
`y'=\sqrt{-x^2-2x+3}+(x+3).(-2x-2)/(2.\sqrt{-x^2-2x+3})(-3<x<1)`
`=\sqrt{-x^2-2x+3}+(x+3).(-x-1)/(\sqrt{-x^2-2x+3})`
`=(-x^2-2x+3+(x+3).(-x-1))/(\sqrt{-x^2-2x+3})`
`=(-2x^2-6x)/(\sqrt{-x^2-2x+3})`
`y'=0<=>-2x^2-6x=0<=>-2x.(x+3)=0`
`<=>`\(\left[ \begin{array}{l}x=0(TM)\\x=-3(Loại)\end{array} \right.\)
Ta có: `y(-3)=0;y(0)=3\sqrt{3};y(1)=0`
Vậy `max\text( )y=y(0)=3\sqrt{3}`
`min\text ( ) y =y(-3)=y(1)=0`
