Bài 1.
`a)`\(\sqrt{3x-1}=5\)
\(ĐK:x\ge\dfrac{1}{3}\)
\(\Leftrightarrow3x-1=25\)
\(\Leftrightarrow3x=26\)
\(\Leftrightarrow x=\dfrac{26}{3}\left(tm\right)\)
`b)`\(\sqrt{\left(2x-3\right)^2}=7\)
\(ĐK:x\in R\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=7;ĐK:x\ge\dfrac{3}{2}\\2x-3=-7:ĐK:x< \dfrac{3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{5;-2\right\}\)
`c)`\(\sqrt{x^2-5x+9}=3\)
\(x^2-5x+9=\left(x-\dfrac{5}{2}\right)^2+\dfrac{11}{4}>0\)
\(ĐK:x\in R\)
\(\Leftrightarrow x^2-5x+9=9\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy \(S=\left\{0;5\right\}\)
