\(A=y^2+\left|x+\dfrac{2}{3}\right|-2\ge-2\\ A_{min}=-2\Leftrightarrow\left\{{}\begin{matrix}y=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=0\end{matrix}\right.\)
Ta thấy: y2 ≥ 0, \(\left|x+\dfrac{2}{3}\right|\ge0\Rightarrow y^2+\left|x+\dfrac{2}{3}\right|-2\ge-2\Rightarrow A\ge-2\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}y^2=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy Amin = - 2 \(\Leftrightarrow\left(x,y\right)=\left(-\dfrac{2}{3};0\right)\)
