a) \(\left\{{}\begin{matrix}x\ge1\\x+y\le3\end{matrix}\right.\)
\(\Rightarrow1\le x\le3\)
\(\Rightarrow Min\left(x^2+y^2\right)\) tại \(x=y=1\left(x+y\le3\right)\)
\(\Rightarrow Min\left(x^2+y^2\right)=1^2+1^2=2\left(x=y=1\right)\)
b) Vi \(1\le x\le3\)
\(\Rightarrow Min\left(x^2+y^2-xy\right)\) tại \(x=y=1\left(x+y\le3\right)\)
\(\Rightarrow Min\left(x^2+y^2-xy\right)=1^2+1^2-1.1=1\left(x=y=1\right)\)
a.
Đặt \(A=x^2+y^2\ge x^2\ge1^2=1\)
\(A_{min}=1\) khi \(\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
b.
Đặt \(B=x^2+y^2-xy=\left(\dfrac{x^2}{4}-xy+y^2\right)+\dfrac{3}{4}x^2=\left(\dfrac{x}{2}-y\right)^2+\dfrac{3}{4}x^2\ge\dfrac{3}{4}x^2\ge\dfrac{3}{4}.1^2=\dfrac{3}{4}\)
\(B_{min}=\dfrac{3}{4}\) khi \(\left\{{}\begin{matrix}x=1\\\dfrac{x}{2}-y=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\end{matrix}\right.\)
