A=\(x^2\)+\(10y^2\)-6xy+6x-22y+2033
=\(x^2-6\text{x}y+6\text{x}+10y^2-22y+2033\)
=\(x^2-6\text{x}\left(y-1\right)+\text{[}3\left(y-1\right)\text{]}^2-\text{[}3\left(y-1\right)\text{]}^2+10y^2-22y+2033\)
=\(\text{[}x-3\left(y-1\right)\text{]}^2-9\left(y^2-2y+1\right)+10y^2-22y+2033\)
=\(\text{[}x-3\left(y-1\right)\text{]}^2+y^2-4y+2024\)
=\(\text{[}x-3\left(y-1\right)\text{]}^2+y^2-4y+4+2020\)
=\(\text{[}x-3\left(y-1\right)\text{]}^2+\left(y-2\right)^2+2020\)
ta có \(\text{[}x-3\left(y-1\right)\text{]}^2\ge0v\text{à}\left(y-2\right)^2\ge0\) và 2020>0
vậy min A=2020 tại \(\left\{{}\begin{matrix}y=2\\x=3\end{matrix}\right.\)
