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Ta có: P= \(5x^2+4xy+y^2+6x+2y+2016\)
= \(\left(4x^2+y^2+1+4x+2y+4xy\right)+\left(x^2+2x+1\right)+2014\)
= \(\left(2x+y+1\right)^2+\left(x+1\right)^2+2014\ge2014\)
(Vì \(\left(2x+y+1\right)^2\ge0;\left(x+1\right)^2\ge0\))
Dấu = khi \(\hept{\begin{cases}2x+y+1=0\\x+1=0\end{cases}< =>}\hept{\begin{cases}y=1\\x=-1\end{cases}}\)
Vậy min P =2014 khi x=-1; y=1
1)
\(2x^2-2xy+5y^2-2x-2y+1=0.\)
\(\Leftrightarrow\left(x^2+y^2+1+2xy-2x-2y\right)+\left(x^2-4xy+4y^2\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)^2+\left(2y-x\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+y-1=0\\2y-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=1\\2y-x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{1}{3}\\x=\frac{2}{3}\end{cases}}}\)
\(A=x^2+4xy+4y^2+2x+4y+1+y^2-4y+4+7\)
=\(\left(x+2y\right)^2+2\left(x+2y\right)+1+\left(y-2\right)^2+7\)
=\(\left(x+2y+1\right)^2+\left(y-2\right)^2+7\ge7\)
vậy \(MinA=7\)Tại \(\hept{\begin{cases}x+2y+1=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=2\end{cases}}\)
\(a,A=3x^2-5x+1\)
\(=3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)-\dfrac{13}{12}\)
\(=3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\)
Với mọi giá trị của x ta có:
\(\left(x-\dfrac{5}{6}\right)^2\ge0\)
\(\Rightarrow3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\ge-\dfrac{13}{12}\)
Vậy Min \(A=-\dfrac{13}{12}\)
Để \(A=-\dfrac{13}{12}\) thì \(x-\dfrac{5}{6}=0\Rightarrow x=\dfrac{5}{6}\)
\(b,B=2x^2+5y^2-4x+2y+4xy+2017\)
\(=\left(2x^2-4x+4xy\right)+5y^2+2y+2017\)
\(=2\left(x^2-2x+2xy\right)+5y^2+2y+2017\)
\(=2\left[x^2-2x\left(1-y\right)+\left(1-y\right)^2\right]+5y^2+2y+2017+2\left(1-y\right)^2\)\(=2\left(x-1+y\right)^2+5y^2+2y+2017-2\left(1-y\right)^2\)
\(=2\left(x+y-1\right)^2+5y^2+2y+2017-2+4y-2y^2\)\(=2\left(x+y-1\right)^2+3y^2+6y+2015\)
\(=2\left(x+y-1\right)^2+3\left(y^2+2y+1\right)+2012\)
\(=2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\)
Với mọi giá trị của x ta có:
\(2\left(x+y-1\right)^2\ge0;3\left(y+1\right)^2\ge0\)
\(\Rightarrow2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\ge2012\) Vậy : Min B = 2012
Để B = 2012 thì \(\left\{{}\begin{matrix}x+y-1=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)