Question

Tìm min H với H=  x² + xy + y² - 3x - 3y
Mình cần 3 cách làm nhé

Answer 1

\(H=x^2+xy+y^2-3x-3y\)

\(=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(xy-x-y+1\right)-3\)

\(=\left(x-1\right)^2+\left(y-1\right)^2+\left(x-1\right)\left(y-1\right)-3\)

\(=\left[\left(x-1\right)^2+2.\frac{1}{2}.\left(x-1\right)\left(y-1\right)+\frac{1}{4}\left(y-1\right)^2\right]+\frac{3}{4}\left(y-1\right)^2-3\)

\(=\left[\left(x-1\right)+\frac{1}{2}\left(y-1\right)\right]^2+\frac{3}{4}\left(y-1\right)^2-3\)

Vì \(\left[\left(x-1\right)+\frac{1}{2}\left(y-1\right)\right]^2+\frac{3}{4}\left(y-1\right)^2\ge0\forall x;y\)

\(\Rightarrow H=\left[\left(x-1\right)+\frac{1}{2}\left(y-1\right)\right]^2+\frac{3}{4}\left(y-1\right)^2-3\ge-3\forall x;y\) có GTNN là - 3

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left[\left(x-1\right)+\frac{1}{2}\left(y-1\right)\right]^2=0\\\frac{3}{4}\left(y-1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}}\)

Vậy \(H_{min}=-3\) tại \(x=1;y=1\)