\(a,\left(x-\frac{1}{2}\right)^2=0\)
=> \(\left|x-\frac{1}{2}\right|=\sqrt{0}=0\)
=> \(x-\frac{1}{2}=0\)
=> \(x=0+\frac{1}{2}=\frac{1}{2}\)
Vậy \(x\in\left\{\frac{1}{2}\right\}\)
\(b,\left(x-\frac{1}{2}\right)^2=\frac{1}{16}\)
=>\(\left|x-\frac{1}{2}\right|=\sqrt{\frac{1}{16}}=\frac{1}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{1}{2}=\frac{1}{4}\\x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\\x=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{4};\frac{1}{4}\right\}\)
\(c,x^5=x^2\)
=> \(x^5-x^2=0\)
=> \(x^2\left(x^3-1\right)=0\)
=> \(\left[{}\begin{matrix}x^2=0\\x^3-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\sqrt{0}=0\\x=\sqrt[3]{0+1}=1\end{matrix}\right.\)
Vậy \(x\in\left\{0;1\right\}\)
