Question

tìm x , y sao cho A = 2x^2 + 9y^2 - 6xy -12y + 2004 có giá trị nhỏ nhất

Answer 1

\(A=2x^2+9y^2-6xy-12y+2004\)

\(=\left(9y^2-6xy-12y\right)+2x^2+2004\)

\(=\left[9y^2-2.3y\left(x+2\right)+\left(x+2\right)^2\right]+2x^2-\left(x+2\right)^2+2004\)\(=\left(3y-x-2\right)^2+2x^2-x^2-4x-4+2004\)

\(=\left(3y-x-2\right)^2+x^2-4x+4+1996\)

\(=\left(3y-x-2\right)^2+\left(x-2\right)^2+1996\)

Với mọi giá trị của x;y ta có:

\(\left(3y-x-2\right)^2\ge0;\left(x-2\right)^2\ge0\)

\(\Rightarrow\left(3y-x-2\right)^2+\left(x-2\right)^2+1996\ge1996\)

Vậy Min A = 1996

Để A = 1996 thì \(\left\{{}\begin{matrix}3y-x-2=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3y-4=0\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=\dfrac{4}{3}\\x=2\end{matrix}\right.\)