a/ Đặt A=\(x^2+3\cdot\left|y-2\right|-1\)

Vì \(\left\{{}\begin{matrix}x^2\ge0\forall x\\3\cdot\left|y-2\right|\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow x^2+3\cdot\left|y-2\right|\ge0\)

\(\Rightarrow x^2+3\cdot\left|y-2\right|-1\ge-1\)

Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

Vậy \(A_{MIN}=-1\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

b/ Đặt \(B=x+\left|x\right|\)

Vì \(\left|x\right|\ge0\forall x\Rightarrow x+\left|x\right|\ge0\)

Dấu ''='' xảy ra \(\Leftrightarrow x\le0\)

Vậy \(B_{MIN}=0\) khi \(x\le0\)