2.
a) \(\left|2x-5\right|=x-1\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=x-1\\2x-5=1-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-x=\left(-1\right)+5\\2x+x=1+5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}1x=4\\3x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4:1\\x=6:3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{4;2\right\}.\)
b) \(\left(x-2013\right)^{2014}=1\)
\(\Rightarrow\left(x-2013\right)^{2014}=\left(\pm1\right)^{2014}\)
\(\Rightarrow x-2013=\pm1.\)
\(\Rightarrow\left[{}\begin{matrix}x-2013=1\\x-2013=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1+2013\\x=\left(-1\right)+2013\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2014\\x=2012\end{matrix}\right.\)
Vậy \(x\in\left\{2014;2012\right\}.\)
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); f(– 2).
