Đáp án + Giải thích các bước giải:
Ta có :
`(2x-1)^{7}=(2x-1)^{5}`
`→(2x-1)^{7}-(2x-1)^{5}=0`
`→(2x-1)^{5}((2x-1)^{2}-1)=0`
`→` \(\left[ \begin{array}{l}(2x-1)^{5}=0\\(2x-1)^{2}-1=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}2x-1=0\\(2x-1)^{2}=1\end{array} \right.\)
`→` \(\left[ \begin{array}{l}2x=1\\2x-1=±1\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=\frac{1}{2}\\2x=2\\2x=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=\frac{1}{2}\\x=1\\x=0\end{array} \right.\)
Vậy `x∈{\frac{1}{2};1;0}`
