\(\left(\frac{-1}{25}\right)^{14}:\left(2x-1\right)^2=\left(\frac{1}{5}\right)^{26}\)
=> (2x-1)2 = \(\left(\frac{-1}{25}\right)^{14}:\left(\frac{1}{5}\right)^{26}\)
=> ( 2x - 1 )2 = \(\left(\frac{-1}{25}\right)^{14}:\left(\frac{1}{25}\right)^{13}\)
=> ( 2x - 1 )2 = \(\left[\left(\frac{-1}{25}\right)^{13}.\left(\frac{-1}{25}\right)\right]:\left(\frac{1}{25}\right)^{13}\)
=> ( 2x - 1 )2 = \(\frac{1}{25}\)
=> ( 2x - 1 )^2 = \(\left(\frac{1}{5}\right)^2\)
=> \(\hept{\begin{cases}2x-1=\frac{1}{5}\\2x-1=\frac{-1}{5}\end{cases}}\)
=> \(\hept{\begin{cases}2x=\frac{1}{5}+1\\2x=\frac{-1}{5}+1\end{cases}}\)
=> \(\hept{\begin{cases}x=\frac{3}{5}\\x=\frac{2}{5}\end{cases}}\)
Vậy x = 3/5 hay x = 2/5
