\(C=x^2+3y^2+2xy+3x+4y+5.\)
\(C=\left(x^2+2xy+y^2\right)+\left(2y^2+4y+2\right)+3\)
\(C=\left(x+y\right)^2+2\left(y^2+2y+1\right)+3\)
\(C=\left(x+y\right)^2+2\left(y+1\right)^2+3\)
Vì \(\left(x+y\right)^2\ge0\) dấu = khi \(x+y=0\Leftrightarrow x=-y\)
\(\left(y+1\right)^2\ge0\) dấu = khi \(y+1=0\Leftrightarrow y=-1\)
\(3>0\)
\(\Rightarrow\left(x+y\right)^2+2\left(y+1\right)^2+3\ge3\) dấu = khi \(x=1;y=-1\)
\(\Rightarrow C=x^2+3y^2+2xy+3x+4y+5\ge3\) dấu = khi \(x=1;y=-1\)
Vậy \(C_{min}=3\) khi \(x=1;y=-1\)
Đúng 0
Bình luận (0)
