\(4x^2-1\ge0\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge\frac{1}{2}\\x\le-\frac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-2\ge0\\x-3\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge3\end{matrix}\right.\) \(\Rightarrow x\ge3\)
a) Để căn thức có nghĩa
\(\Rightarrow4x^2-1\ge0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\2x+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\frac{1}{2}\\x< -\frac{1}{2}\end{matrix}\right.\)
Vậy căn thức trên có nghĩa \(\Leftrightarrow x\ge\frac{1}{2}\) hoặc \(x< -\frac{1}{2}\)
b) Để căn thức có nghĩa
\(\Rightarrow\left\{{}\begin{matrix}x-2\ge0\\x-3\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge3\end{matrix}\right.\)
\(\Leftrightarrow x\ge3\)
Vậy căn thức trên có nghĩa \(\Leftrightarrow x\ge3\)
a/ Để... có nghĩa\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\2x+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\2x+1\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge\frac{1}{2}\\x\le\frac{-1}{2}\end{matrix}\right.\)
b/ Để... có nghĩa
\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x-3\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ge3\end{matrix}\right.\Leftrightarrow x\ge3\)
