Vì : $|x-y|$ $≥$ $0$ $∀$ $x;y$
$|y+\dfrac{9}{25}|$ $≥$ $0$ $∀$ $y$
Mà $|x-y| + |y + \dfrac{9}{25}| = 0$
$⇒$ $\left \{ {{x-y=0} \atop {y+\dfrac{9}{25}=0}} \right.$
$⇔$ $\left \{ {{x-y=0} \atop {y=-\dfrac{9}{25}}} \right.$
$⇔$ $\left \{ {{x-(-\dfrac{9}{25})=0} \atop {y=-\dfrac{9}{25}}} \right.$
$⇔$ $\left \{ {{x+\dfrac{9}{25}=0} \atop {y=-\dfrac{9}{25}}} \right.$
$⇔$ $\left \{ {{x=-\dfrac{9}{25}} \atop {y=-\dfrac{9}{25}}} \right.$
Vậy $x=y=\dfrac{-9}{25}$.
