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\(K=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2012=\left(x-y\right)^2+\left(y-2\right)^2+2012\ge2012\)Min K = 2012 <=> x = y = 2
\(A=x^2+2y^2+2xy+2x-4y+2016\)
\(=x^2+y^2+y^2+2xy+2x+2y-6y+2016\)
\(=\left(x^2+2xy+y^2\right)+\left(y^2-6y+9\right)+\left(2x+2y\right)+2007\)
\(=\left(x+y\right)^2+\left(y-3\right)^2+2\left(x+y\right)+2007\)
\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2006\)
Vì \(\hept{\begin{cases}\left(x+y+1\right)^2\ge0;\forall x,y\\\left(y-3\right)^2\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(x+y+1\right)^2+\left(y-3\right)^2\ge0;\forall x,y\)
\(\Rightarrow\left(x+y+1\right)^2+\left(y-3\right)^2+2006\ge0+2006;\forall x,y\)
Hay \(A\ge2006;\forall x,y\)
Dấu"=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)
Vậy \(A_{min}=2006\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)
a: A=x^2-2xy+y^2+y^2-4y+4+1
=(x-y)^2+(y-2)^2+1>=1
Dấu = xảy ra khi x=y=2
b: B=4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1-2
=(2x+2y)^2+(x-1)^2+(y+1)^2-2>=-2
Dấu = xảy ra khi x=1 và y=-1
\(A=\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(y^2-6y+9\right)+2006\)\(=\left(x+y\right)^2+2\left(x+y\right)+1+\left(y-3\right)^2+2006\)
\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2006\)
Ta có: \(\left(x+y+1\right)^2+\left(y-3\right)^2\ge0\left(\forall x;y\right)\)
\(\Rightarrow A\ge2006\).
Vậy MIN A = 2006 \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)
a ) \(x^2-x+1\)
\(\Leftrightarrow\left(x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right)+\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có : \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN là \(\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}.\)
a)\(M=x^2-2xy+2y^2-4y+2016\)
\(=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2012\)
\(=\left(x-y\right)^2+\left(y-2\right)^2+2012\ge2012\)
Dấu = khi \(\begin{cases}\left(x-y\right)^2=0\\\left(y-2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y=0\\y-2=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x=y\\y=2\end{cases}\)\(\Leftrightarrow x=y=2\)
Vậy MinM=2012 khi x=y=2
b)\(N=x^2-2xy+2x+2y^2-4y+2016\)
\(=\left(x^2-2xy+2x+y^2-2y+1\right)+\left(y^2-2y+1\right)+2014\)
\(=\left(x-y+1\right)^2+\left(y-1\right)^2+2014\ge2014\)
Dấu = khi \(\begin{cases}\left(x-y+1\right)^2=0\\\left(y-1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y+1=0\\y-1=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x-y+1=0\\y=1\end{cases}\)\(\Leftrightarrow\begin{cases}x-1+1=0\\y=1\end{cases}\)\(\Leftrightarrow\begin{cases}x=0\\y=1\end{cases}\)
Vậy MinN=2014 khi x=0;y=1