Đáp án:
`x∈{1;1/2}`
Giải thích các bước giải:
`(2x-1)^2018=(2x-1)^2019`
`->(2x-1)^2019-(2x-1)^2018=0`
`->(2x-1)^2018 . (2x-1-1)=0`
`->(2x-1)^2018 . (2x-2)=0`
`->`\(\left[ \begin{array}{l}(2x-1)^{2018}=0\\2x-2=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}2x-1=0\\x-1=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=1\end{array} \right.\)
