a) Ta có \(x^2\ge0;\left(y-\dfrac{1}{3}\right)^2\ge0\)
\(x^2+\left(y-\dfrac{1}{3}\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có \(\left(\dfrac{1}{2}x-3\right)^4\ge0;\left(y^2-\dfrac{1}{4}\right)^6\ge0\)
\(\left(\dfrac{1}{2}x-3\right)^4+\left(y^2-\dfrac{1}{4}\right)^6\le0\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-3=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=6\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
