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$\Big(\dfrac{1}{2}x-5\Big)^4≥0$

$\Big(y^2-\dfrac{1}{4}\Big)^2≥0$

$→\Big(\dfrac{1}{2}x-5\Big)^4+\Big(y^2-\dfrac{1}{4}\Big)^2≥0$

mà theo đề bài: $\Big(\dfrac{1}{2}x-5\Big)^4+\Big(y^2-\dfrac{1}{4}\Big)^2≤0$

$→x,y$ chỉ xuất hiện khi $\Big(\dfrac{1}{2}x-5\Big)^4+\Big(y^2-\dfrac{1}{4}\Big)^2=0$ để thỏa mãn đề bài

$→\begin{cases}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{cases}$

$→\begin{cases}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{cases}$

$→\begin{cases}x=10\\y=\pm\dfrac{1}{2}\end{cases}$