Ta có A = 5x2 - 2xy + 2y2 - 4x + 2y + 3
=> 2A = 10x2 - 4xy + 4y2 - 8x + 4y + 6
= (x2 - 4xy + 4y2) - 2(x - 2y) + 1 + 9x2 - 6x + 1 + 4
= \(\left(x-2y\right)^2-2\left(x-2y\right)+1+9\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+4\)
\(=\left(x-2y-1\right)^2+9\left(x-\frac{1}{3}\right)^2+4\)\(\ge4\)
=> A \(\ge\)2
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-2y-=0\\x-\frac{1}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2y=1\\x=\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{3}\\x=\frac{1}{3}\end{cases}}\)
Vậy khi x = 1/3 ; y = -1/3 thì A đạt GTNN
\(A=5x^2+2y^2-2xy-4x+2y\)\(+3\)
\(=\left(x^2-2xy+y^2\right)+\)\(\left(4x^2-4x+1\right)+\)\(\left(y^2+2y+1\right)+1\)
\(Tacó\)
\(A=5x^2+2y^2-2xy-4x+2y+3\)
\(\Rightarrow2A=10x^2+4y^2-4xy-8x+4y+6\)
\(=\left(x^2-4xy+4y^2\right)-\left(2x-4y\right)+1+\left(9x^2-6x+1\right)+4\)
\(=\left(x-2y\right)^2-2\left(x-2y\right)+1+\left(3x-1\right)^2+4\)
\(=\left(x-2y-1\right)^2+\left(3x-1\right)^2+4\)
Vì \(\left(x-2y-1\right)^2\ge0\), \(\left(3x-1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-2y-1\right)^2+\left(3x-1\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-2y-1\right)^2+\left(3x-1\right)^2+4\ge4\forall x,y\)
\(\Rightarrow2A\ge4\)\(\Rightarrow A\ge2\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2y-1=0\\3x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2y=x-1\\x=\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}2y=\frac{-2}{3}\\x=\frac{1}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}y=\frac{-1}{3}\\x=\frac{1}{3}\end{cases}}\)
Vậy \(minA=2\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{3}\end{cases}}\)
\(A=5x^2+2y^2-2xy-4x\)\(+2y+3\)
\(=\left(x^2-2xy+y^2\right)+\left(4x^2-4x+1\right)\)\(+\left(y^2+2y+1\right)+1\)
\(=\left(x-y\right)^2+\left(2x-1\right)^2+\left(y+1\right)^2\)\(+1\)
\(Tacó\)\(\left(x-y\right)^2\ge0\)\(\forall x,y\in R\)
\(\left(2x-1\right)^2\ge0\)\(\forall x\in R\)
\(\left(y+1\right)^2\ge0\forall y\in R\)
\(\Rightarrow\)\(\left(x-y\right)^2+\left(2x-1\right)^2+\left(y+1\right)^2+1\)\(\ge1\)