a) |2,5 - x| = 1,3 |2,5 - x| = 1,3
=> |2,5 - x| = |2,5 - x| = 1,3 : 1,3
=> |2,5 - x| = |2,5 - x| = 1
=> \(\left[{}\begin{matrix}2,5-x=1\\2,5-x=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1,5\\x=3,5\end{matrix}\right.\)
Vậy x = 1,5 hoặc x = 3,5
b) 1,6 - |x - 0,2| = 0
=> |x - 0,2| = 1,6
=> \(\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)
Vậy x = 1,8 hoặc x = -1,4
c) \(\left(x-\dfrac{1}{2}\right)^2\) = 0
=> \(x-\dfrac{1}{2}=0\)
=> \(x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
d) (x - 2)2 = 1
=> \(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy x = 3 hoặc x = 1
e) (2x - 1)3 = -8
=> (2x - 1)3 = (-2)3
=> 2x - 1 = -2
=> 2x = -1
=> x = \(\dfrac{-1}{2}\)
Vậy \(x=\dfrac{-1}{2}\)
f) \(\left(x+\dfrac{1}{2}\right)^2\) = \(\dfrac{1}{16}\)
=> \(\left(x+\dfrac{1}{2}\right)^2\) = \(\left(\pm\dfrac{1}{4}\right)^2\)
=> \(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{2}\\x=\dfrac{-1}{4}-\dfrac{1}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
Vậy x = \(\dfrac{-1}{4}\) và x = \(\dfrac{-3}{4}\)