a) \(\left(2x-\dfrac{1}{2}\right)^3-15=12\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^3=27\\ \Rightarrow2x-\dfrac{1}{2}=3\\ \Rightarrow x=\dfrac{7}{4}\)
b) \(45-\left(1-\dfrac{x}{3}\right)^2=-4\\ \Rightarrow\left(1-\dfrac{x}{3}\right)^2=49\\ \Rightarrow\left[{}\begin{matrix}1-\dfrac{x}{3}=-7\\1-\dfrac{x}{3}=7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=24\\x=-18\end{matrix}\right.\)
a: Ta có: \(\left(2x-\dfrac{1}{2}\right)^3-15=12\)
\(\Leftrightarrow\left(2x-\dfrac{1}{2}\right)^3=27\)
\(\Leftrightarrow2x-\dfrac{1}{2}=3\)
\(\Leftrightarrow2x=\dfrac{7}{2}\)
hay \(x=\dfrac{7}{4}\)
b: Ta có: \(45-\left(1-\dfrac{x}{3}\right)^2=-4\)
\(\Leftrightarrow\left(\dfrac{1}{3}x-1\right)^2=49\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x-1=7\\\dfrac{1}{3}x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=8\\\dfrac{1}{3}x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-18\end{matrix}\right.\)
