`a) sqrt { ( 2x - 1 )^2 } - 5 = 0`
`<=> | 2x - 1 | = 5`
`<=>` \(\left[ \begin{array}{l}2x-1=5\\2x-1=-5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-4\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
Vậy ` x ∈ { -2 ; 3 }`