\(a,\left(x-3^2\right)^3=\dfrac{1}{64}\)
\(x^2-\left(3^2\right)^3=\dfrac{1}{64}\)
\(x^2-3^6=\dfrac{1}{64}\)
\(x^2=\dfrac{1}{64}+3^6\)
\(x^2=729\dfrac{1}{64}\)
\(x=\sqrt{729\dfrac{1}{64}}\approx27\) hoặc \(x\approx-27\)
Vậy \(x\approx27\) hoặc \(x\approx-27\)
\(b,\left(2x-3\right)^2-16=20\)
\(\left(2x-3\right)^2=20+16\)
\(\left(2x-3\right)^2=36\)
\(\left(2x-3\right)^2=6^2\) hoặc \(\left(2x-3\right)^2=-6\)
\(\Rightarrow\left\{{}\begin{matrix}2x+3=-6\\2x+3=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{9}{2};\dfrac{3}{2}\right\}\)
\(\left(x-\dfrac{3}{2}\right)^3=\dfrac{1}{64}\\ \left(x-\dfrac{3}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\\ =>\left\{{}x-\dfrac{3}{2}=\dfrac{1}{4}}\left\{{}x=\dfrac{1}{4}+\dfrac{3}{2}}\left\{{}x=\dfrac{7}{4}}\)
