Question

tìm min A=x²-4xy+5y²+10x-22y+2006

Answer 1

\(A=x^2-4xy+5y^2+10x-22y+2006\)

\(=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25+y^2-2y+1+1980\)   

\(=\left(x-2y\right)^2+2.\left(x-2y\right).5+5^2+\left(y-1\right)^2+1980\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2\ge1980\forall x;y\)

Dấu "=" xảy ra khi:

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy Min A = 1980 khi x = -3 và y = 1