\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)
\(\Leftrightarrow x+\frac{1}{2}=\pm\frac{2}{5}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{2}{5}\\x+\frac{1}{2}=-\frac{2}{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{10}\\x=-\frac{9}{10}\end{cases}}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{2}{5}\right)^2\)
\(x+\frac{1}{2}=\pm\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=-\frac{2}{5}\\x+\frac{1}{2}=\frac{2}{5}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{2}{5}-\frac{1}{2}=-\frac{9}{10}\\x=\frac{2}{5}-\frac{1}{2}=-\frac{1}{10}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-\frac{9}{10}\text{ ; }-\frac{1}{10}\right\}\)
Sosuke đến dòng thứ hai thì bạn thêm \(\pm\) luôn nha !
\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{2}{5}\right)^2\)
\(x+\frac{1}{2}=\pm\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=-\frac{2}{5}\\x+\frac{1}{2}=\frac{2}{5}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{2}{5}-\frac{1}{2}=-\frac{9}{10}\\x=\frac{2}{5}-\frac{1}{2}=-\frac{1}{10}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-\frac{9}{10}\text{ ; }-\frac{1}{10}\right\}\)
\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2=\left(\frac{-2}{5}\right)^2\)
\(x+\frac{1}{2}=\frac{2}{5}\) \(x+\frac{1}{2}=-\frac{2}{5}\)
\(x=\frac{2}{5}-\frac{1}{2}\) \(x=-\frac{2}{5}-\frac{1}{2}\)
\(x=\frac{4}{10}-\frac{5}{10}\) \(x=-\frac{4}{10}-\frac{5}{10}\)
\(x=-\frac{1}{10}\) \(x=-\frac{9}{10}\)
Vậy \(x\in\left\{-\frac{1}{10};-\frac{9}{10}\right\}\)
