`f(x)=\sqrt{log_{1/ 2} \ {3-2x-x^2}/{x+1}`
`ĐK: x+1\ne 0<=>x\ne -1`
`\qquad log_{1/ 2} \ {3-2x-x^2}/{x+1}\ge 0`
`<=>{3-2x-x^2}/{x+1}\ge 1`
`<=>{3-2x-x^2}/{x+1}-1\ge 0`
`<=>{3-2x-x^2-(x+1)}/{x+1}\ge 0`
`<=>{-x^2-3x+2}/{x+1}\ge 0`
$⇔\left[\begin{array}{l}\begin{cases}-x^2-3x+2\ge 0\\x+1>0\end{cases}\\\begin{cases}-x^2-3x+2\le 0\\x+1<0\end{cases}\end{array}\right.$
$⇔\left[\begin{array}{l}\begin{cases}\dfrac{-3-\sqrt{17}}{2}\le x\le \dfrac{-3+\sqrt{17}}{2}\\x>-1\end{cases}\\\begin{cases}x\ge \dfrac{-3+\sqrt{17}}{2}\ hoặc \ x\le \dfrac{-3-\sqrt{17}}{2}\\x<-1\end{cases}\end{array}\right.$
$⇔\left[\begin{array}{l}-1<x\le \dfrac{-3+\sqrt{17}}{2}\\x\le \dfrac{-3-\sqrt{17}}{2}\end{array}\right.$
Vậy
`TXĐ`: `D=(-∞;{-3-\sqrt{17}}/2]∪(-1;{-3+\sqrt{17}}/2]`
