\(\left(2x-1\right)^4=16=\left(\pm2\right)^4\\ =>\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\left(2x+1\right)^3=125=5^3\\ =>2x+1=1\\ =>x=2\)
\(\left(2x-1\right)^4=16\)
\(\left(2x-1\right)^4=2^4\)
\(2x-1=2\)
\(2x=3\)
\(x=\dfrac{3}{2}\)
\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
(2\(x\) - 1)4 = 16 (2\(x\) + 1)3 = 125
(2\(x\) - 1)4 = 24 (2\(x\) +1)3 = 53
\(\left[{}\begin{matrix}2x-1=-2\\2x-1=2\end{matrix}\right.\) 2\(x\) + 1 = 5
\(\left[{}\begin{matrix}2x=-1\\2x=3\end{matrix}\right.\) 2\(x\) = 4
\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\) \(x\) = 2
Vậy \(x\) \(\in\) { - \(\dfrac{1}{2}\); \(\dfrac{3}{2}\)} vậy \(x\) = 2
