$(2x+\dfrac 3 5)^2-\dfrac{24}{25}=1\\\Leftrightarrow (2x+\dfrac{3}{5})^2=\dfrac{49}{25}\\\Leftrightarrow \left[\begin{array}{1}2x+\dfrac{3}{5}=\dfrac{7}{5}\\2x+\dfrac{3}{5}=-\dfrac{7}{5}\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}2x=\dfrac{4}{5}\\2x=-2\end{array}\right.\\\Leftrightarrow \left[\begin{array}{1}x=\dfrac{2}{5}\\x=-1\end{array}\right.$
Vậy $x=\dfrac{2}{5},x=-1$
GIải
\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{24}{25}=1\)
\(\left(2x+\dfrac{3}{5}\right)^2\) \(=1+\dfrac{24}{25}\)
\(\left(2x+\dfrac{3}{5}\right)^2\) \(=\dfrac{49}{25}\)
\(4x+\dfrac{9}{25}\) \(=\dfrac{49}{25}\)
\(4x\) \(=\dfrac{49}{25}-\dfrac{9}{25}\)
\(4x\) \(=\dfrac{8}{5}\)
\(x\) \(=4:\dfrac{8}{5}\)
\(x\) \(=\dfrac{5}{2}\)
