a/ \(\sqrt{3x-1}=4\)
\(3x-1=16\)
\(3x=17\)
\(x=\dfrac{17}{3}\)
b/ \(\sqrt{25x^2-10x+1}=9\)
\(\sqrt{\left(5x-1\right)^2=9}\)
\(5x-1=9\)
\(5x=10\)
\(x=2\)
\(a,ĐK:x\ge\dfrac{1}{3}\\ PT\Leftrightarrow3x-1=16\Leftrightarrow x=\dfrac{17}{3}\left(tm\right)\\ b,ĐK:x\in R\\ PT\Leftrightarrow\left|5x-1\right|=9\\ \Leftrightarrow\left[{}\begin{matrix}5x-1=9\left(x\ge\dfrac{1}{5}\right)\\5x-1=-9\left(x< \dfrac{1}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-\dfrac{8}{5}\left(tm\right)\end{matrix}\right.\\ c,ĐK:x\ge2\\ PT\Leftrightarrow2\cdot4\sqrt{x-2}-\dfrac{1}{2}\cdot2\sqrt{x-2}+5\sqrt{x-2}=24\\ \Leftrightarrow12\sqrt{x-2}=24\Leftrightarrow\sqrt{x-2}=2\\ \Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
b: Ta có: \(\sqrt{25x^2-10x+1}=9\)
\(\Leftrightarrow\left|5x-1\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=9\\5x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=10\\5x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{8}{5}\end{matrix}\right.\)
