Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2\right|+\left|y+3\right|=\left|x-2\right|+\left|-y-3\right|\)
\(\ge\left|x-2-y-3\right|=\left|x-y-5\right|=3\)
Dấu "=" xảy ra khi \(x=y+2\)
Vậy với \(x=y+2\) thì \(A_{Min}=3\)
\(\left\{{}\begin{matrix}x-y=2\\A=\left|x-2\right|+\left|y+3\right|\end{matrix}\right.\)\(\Leftrightarrow A=\left|y\right|+\left|y+3\right|\ge\left|y-y-3\right|=\left|-3\right|=3\)
Đẳng thức khi \(\left\{{}\begin{matrix}x-2=y\\y\left(y+3\right)\le0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-y=2\\\left\{{}\begin{matrix}-3\le y\le0\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)
