\(\left(-\dfrac{1}{25}\right)^{14}:\left(2x-1\right)^2=\left(\dfrac{1}{5}\right)^{26}\)
\(\Rightarrow\left(2x-1\right)^2=\left(-\dfrac{1}{25}\right)^{14}:\left(\dfrac{1}{5}\right)^{26}\)
\(\Rightarrow\left(2x-1\right)^2=625\)
\(\Rightarrow\left(2x-1\right)^2=25^2\)
\(\Rightarrow TH1:2x-1=25\)
\(\Rightarrow x=\dfrac{25+1}{2}\)
\(\Rightarrow x=13\)
\(\Rightarrow TH2:2x-1=\left(-25\right)\)
\(\Rightarrow x=\dfrac{\left(-25\right)+1}{2}\)
\(\Rightarrow x=\left(-12\right)\)
Vậy \(x\in\left\{-12;13\right\}\)