a) \(\sqrt{4x^2-16}\)
\(=\)\(\sqrt{\left(2x\right)^2-4^2}\)
\(=\sqrt{\left(2x+4\right)\left(2x-4\right)}\)
để phương trình trên có nghĩa
⇒2x-4≥0
⇒x≥2
a) \(ĐK:4x^2-16\ge0\)
\(\Leftrightarrow4x^2\ge16\Leftrightarrow x^2\ge4\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
b) \(ĐK:9x^2-25\ge0\)
\(\Leftrightarrow9x^2\ge25\)\(\Leftrightarrow x^2\ge\dfrac{25}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{3}\\x\le-\dfrac{5}{3}\end{matrix}\right.\)
\(a,ĐK:4x^2-16\ge0\Leftrightarrow4\left(x-2\right)\left(x+2\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\ge0\\x+2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
\(b,ĐK:9x^2-25\ge0\Leftrightarrow\left(3x-5\right)\left(3x+5\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-5\ge0\\3x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x-5\le0\\3x+5\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{3}\\x\le-\dfrac{5}{3}\end{matrix}\right.\)
