\(\left\{{}\begin{matrix}16-x^2\ge0\\2x+1>0\\x^2-8x+14\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4\le x\le4\\x>-\dfrac{1}{2}\\\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}< x\le4-\sqrt{2}\)
xác định \(< =>\left\{{}\begin{matrix}\sqrt{16-x^2}\ge0\\\sqrt{2x+1}>0\\\sqrt{x^2-8x+14}\ge0\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}-4\le x\le4\\x>-\dfrac{1}{2}\\\left[{}\begin{matrix}x\le4-\sqrt{2}\\x\ge4_{ }+\sqrt{2}\end{matrix}\right.\\\end{matrix}\right.\)\(< =>-\dfrac{1}{2}< x\le4-\sqrt{2}\)
ĐKXĐ: \(\left\{{}\begin{matrix}16-x^2\ge0\\2x+1>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2\le16\\x>-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le-4\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge4\\x>-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x\ge4\)
