Tìm x biết (2x-1)^7=(2x-1)^5

Giải:

`(2x - 1)^7 = (2x - 1)^5`

`-> (2x - 1)^7 - (2x - 1)^5 = 0`

`-> (2x - 1)^5 . [(2x - 1)^2 - 1] = 0`

`->` \(\left[ \begin{array}{l}(2x - 1)^5 = 0\\(2x - 1)^2 - 1 = 0\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}(2x - 1)^5 = 0\\(2x - 1)^2 - 1 = 0\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}2x - 1 = 0\\(2x - 1)^2 = 1\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}2x = 0 + 1 = 1\\2x - 1 = ±1\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\\left[ \begin{array}{l}2x = 2\\2x = 0\end{array} \right.\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\\left[ \begin{array}{l}x = 1\\x = 0\end{array} \right.\end{array} \right.\) 

Vậy `x ∈ { 1/2 ; 0 ; 1 }`