2b) Ta có:
\(\left(x+\dfrac{3}{5}\right)^2=\dfrac{4}{25}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{5}=\dfrac{2}{5}\\x+\dfrac{3}{5}=\dfrac{-2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=-1\end{matrix}\right.\)
2.b) Ta có: \(\left(x+\dfrac{3}{5}\right)^2=\dfrac{4}{25}\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x+\dfrac{3}{5}=\dfrac{2}{5}\\x+\dfrac{3}{5}=\dfrac{-2}{5}\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=-1\end{matrix}\right.\)
CHO MÌNH 1 TICK NHA
2b:
Ta có: \(\left(x+\dfrac{3}{5}\right)^2=\dfrac{4}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{5}=\dfrac{2}{5}\\x+\dfrac{3}{5}=-\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-1\end{matrix}\right.\)
