Để $A=0$
$⇒ x^2 + (\dfrac{1}{2}-y)^2=0$
Mà :$x^2$ $≥$ $0$ $∀$ $x$
$(\dfrac{1}{2}-y)^2$ $≥$ $0$ $∀$ $y$
$⇒$ $\left \{ {{x^2=0} \atop {\dfrac{1}{2}-y=0}} \right.$
$⇔$ $\left \{ {{x=0} \atop {y=\dfrac{1}{2}}} \right.$
Vậy `(x;y)=(0;\frac{1}{2})` thì biểu thức $A=0$.