Bài 2:
a: Ta có: \(\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\)
hay x=6
b: ta có: \(\sqrt{36x^2-12x+1}=5\)
\(\Leftrightarrow\left|6x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}6x-1=5\\6x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)