Question

tìm GTNN của biểu thức sau

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

Answer 1

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

\(4C=4x^2+4xy+4y^2-12x-12y\)

\(4C=\left(4x^2+4xy+y^2\right)-6\left(2x+y\right)+3\left(y^2-2y+1\right)-3\)

\(4C=\left(2x+y\right)^2-6\left(2x+y\right)+9+3\left(y-1\right)^2-12\)

\(4C=\left(2x+y-3\right)^2+3\left(y-1\right)^2-12\)

\(C=\dfrac{\left(2x+y-3\right)^2+3\left(y-1\right)^2}{4}-3\ge-3\forall x,y\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}2x+y-3=0\\y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

\(MinC=-3\Leftrightarrow x=y=1\)