\(a,\sqrt{x^2-9}=\sqrt{\left(x-3\right)\left(x+3\right)}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+3\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+3\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\ge-3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\le-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\le-3\\x\ge3\end{matrix}\right.\)
\(b,\sqrt{49x^2-24x+4}=\sqrt{\left(7x-2\right)^2}\ge0\forall x\)
\(\Rightarrow\) Căn thức có nghĩa \(\forall x\)
`a,` Điều kiện: `x^2 - 9 >=0 <=>` \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)
`b,` Điều kiện: `49x^2-24x+4 = (7x-2)^2 >=0`.
`-> x in RR`.
