1:
a: Ta có: \(\left(2x-3\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=7\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{2}{3}\\x-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=-\dfrac{5}{12}\end{matrix}\right.\)
2:
a: Ta có: \(\left(x-\dfrac{2}{3}\right)^3=\dfrac{1}{27}\)
\(\Leftrightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
hay x=1
b: Ta có: \(\left(2x-5\right)^4=-81\)
mà \(\left(2x-5\right)^4\ge0\forall x\)
nên \(x\in\varnothing\)
