\(A=\left|x-3\right|+x^2+y^2+1\)
Trước tiên ta xét \(\left|x-3\right|\) có \(2\) trường hợp \(\left[{}\begin{matrix}x>3\\x< 3\end{matrix}\right.\)
Để \(A\) đạt \(Min\) thì \(|x-3|\) đạt \(Min\), vậy sẽ xét trường hợp \(x< 3\)
Khi đó \(\left|x-3\right|=3-x\Leftrightarrow A=3-x+x^2+y^2+1\)
\(\Leftrightarrow A=x^2-x+4+y^2=x^2-x+\dfrac{1}{4}+\dfrac{15}{4}+y^2\)
\(\Leftrightarrow A=\left(x^2-x+\dfrac{1}{4}\right)+y^2+\dfrac{15}{4}\)
\(\Leftrightarrow A=\left(x-\dfrac{1}{2}\right)^2+y^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x,y\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=0\\y^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=0\end{matrix}\right.\)
Vậy với \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=0\end{matrix}\right.\) thì \(A_{Min}=\dfrac{15}{4}\)
\(B=\left|x-100\right|+\left(x-y\right)^2+100\)
Ta thấy: \(\left\{{}\begin{matrix}\left|x-100\right|\ge0\forall x\\\left(x-y\right)^2\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left|x-100\right|+\left(x-y\right)^2\ge0\forall x,y\)
\(\Rightarrow B=\left|x-100\right|+\left(x-y\right)^2+100\ge100\forall x,y\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\left|x-100\right|=0\\\left(x-y\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-100=0\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=100\)
Vậy với \(x=y=100\) thì \(B_{Min}=100\)
