Tìm x (8x-1)^ 2x+1= 5^2x+1

My Nguyễn (Pé sứa cuti)

$\left(8x-1\right)^{2x+1}=5^{2x+1}$

$\left(\frac{8x-1}{5}\right)^{2x-1}=1$

$\left[\begin{matrix}2x-1=0 \\ \frac{8x-1}{5}=0\end{matrix}\right.$

$\left[\begin{matrix}x=\frac{1}{2} \\ 8x=6\end{matrix}\right.$

$\left[\begin{matrix}x=\frac{1}{2} \\ x=\frac{3}{4}\end{matrix}\right.$

Vậy $x\in\left\lbrace\frac{1}{2};\frac{3}{4}\right\rbrace$.