Ta có:
$(x+\dfrac{1}{2})^2 = \dfrac{1}{25}$
$⇒(x+\dfrac{1}{2})^2 = (\dfrac{1}{5})^2$ hoặc $(x+\dfrac{1}{2})^2 = (-\dfrac{1}{5})^2$
$⇒x+\dfrac{1}{2}=\dfrac{1}{5}$ hoặc $⇒x+\dfrac{1}{2}=-\dfrac{1}{5}$
$⇒x=-\dfrac{3}{10}$ hoặc $x=\dfrac{-7}{10}$
Vậy $x=-\dfrac{3}{10}$ hoặc $x=\dfrac{-7}{10}$ thì $(x+\dfrac{1}{2})^2 = \dfrac{1}{25}$.
