Question

Tìm GTNN của bt sau:

 C=x^2+3y^2+2xy+3x+4y+5

 

Answer 1

\(C=x^2+3y^2+2xy+3x+4y+5.\)

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

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

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

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 \(x=1;y=-1\)