\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{25}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{5}\\x+\dfrac{1}{2}=-\dfrac{2}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}-\dfrac{1}{2}\\x=-\dfrac{2}{5}-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)
#\(Toru\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{25}\)
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{2}{5}^2\)\(=\left(-\dfrac{2}{5}\right)^2\)
\(\Rightarrow\)\(x+\dfrac{1}{2}=\dfrac{2}{5}\)\(=-\dfrac{2}{5}\)
\(+>x=\dfrac{2}{5}-\dfrac{1}{2}\)
\(x=-\dfrac{1}{10}\)
\(+>x=-\dfrac{2}{5}-\dfrac{1}{2}\)
\(x=-\dfrac{9}{10}\)
\(\Rightarrow x\in\left\{-\dfrac{1}{10};-\dfrac{9}{10}\right\}\)
