Ta có : \(x-y=2\Rightarrow x=2+y\)
\(\Rightarrow P=\left(2+y\right)y+4\)
\(\Rightarrow P=2y+y^2+4\)
\(\Rightarrow P=y^2+2y+4\)
\(\Rightarrow P=y^2+y+y+1+3\)
\(\Rightarrow P=\left(y^2+y\right)+\left(y+1\right)+3\)
\(\Rightarrow P=y\left(y+1\right)+\left(y+1\right)+3\)
\(\Rightarrow P=\left(y+1\right)\left(y+1\right)+3\)
\(\Rightarrow P=\left(y+1\right)^2+3\)
Ta có : \(\left(y+1\right)^2\ge0\) với \(\forall y\)
\(\Rightarrow\left(y+1\right)^2+3\ge3\) với \(\forall y\)
Dấu"=" xảy ra \(\Leftrightarrow\left(y+1\right)^2=0\Rightarrow y+1=0\Rightarrow y=-1\)
Vậy Min A = 3 \(\Leftrightarrow y=-1\)
Đúng 1
Bình luận (1)
