tim gtnn cua A=x^2+y^2+2xy+2x+2y+3
cho x 0,y 0, x y 2012. a, tim GTLN cua A 2x 2 8xy 2y 2 x 2 2xy y 2 b, tim GTNN cua B 1 2012 x 2 1 2012 y 2
cho x>0,y>0, x+y=2012.
a, tim GTLN cua A= (2x^2+8xy+2y^2)/ (x^2+2xy+y^2)
b, tim GTNN cua B=(1+(2012/x))^2+(1+(2012/y))^2
tim GTNN cua cac don thuc a)x^2 - 4xy + 5y^2 - 2y + 3
b)x^2 - 2xy + 2y^2 - x +y
1) tim GTNN cua cac don thuc a)x^2 - 4xy + 5y^2 - 2y + 3
b)x^2 - 2xy + 2y^2 - x +y
tim GTNN cua cac don thuc a)x^2 - 4xy + 5y^2 - 2y + 3
b)x^2 - 2xy + 2y^2 - x +y
a)Đặt A=\(x^2-4xy+5y^2-2y+3\)
\(\Leftrightarrow x^2-4xy+4y^2+y^2-2y+1+2\)
\(\Leftrightarrow\left(x-2y\right)^2+\left(y-1\right)^2+2\)
Vì \(\left(x-2y\right)^2\ge0;\left(y-1\right)^2\ge0\)
Nên \(\left(x-2y\right)^2+\left(y-1\right)^2+2\ge2\)
Dấu = xảy ra khi \(\hept{\begin{cases}x-2y=0\\y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2y\\y=1\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)
Vậy Min A = 2 khi x = 2 ; y = 1
b)k ko hỉu
a)A= \(x^2-4xy+5y^2-2y+3\)
\(=x^2-4xy+4y^2+y^2-2y+1-2\)
\(=\left(x-2y\right)^2+\left(y-1\right)^2-2\ge-2\)
MIN A=-2 khi\(\orbr{\begin{cases}x-2y=0\\y-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\y=1\end{cases}}}\)Vậy.......
b)\(B=x^2-2xy+2y^2-x+y\)????
Tim GTNN cua
A=\(x^2-2xy+2y^2+2x-10y+2033\)
\(A=x^2-2xy+2y^2+2x-10y+2033\\ =x^2-2xy+y^2+y^2+2x-8y-2y+1+16+2016\\ =\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+1+\left(y^2-8y+16\right)+2016\\ =\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-4\right)^2+2016\\ =\left[\left(x-y\right)^2+2\left(x-y\right)+1\right]+\left(y-4\right)^2+2016\\ =\left(x-y+1\right)^2+\left(y-4\right)^2+2016\\ Do\text{ }\left(y-4\right)^2\ge0\forall y\\ \left(x-y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-y+1\right)^2+\left(y-4\right)^2\ge0\forall x;y\\ \Rightarrow A=\left(x-y+1\right)^2+\left(y-4\right)^2+2016\ge2016\forall x;y\\ Dấu\text{ }''=''\text{ }xảy\text{ }ra\text{ }khi:\left\{{}\begin{matrix}\left(y-4\right)^2=0\\\left(x-y+1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y-4=0\\x-y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x-4+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\\ Vậy\text{ }A_{\left(Min\right)}=2016\text{ }khi\text{ }\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
tim gtnn cua c=2x^2-2xy+2y^2+4y-1
cho x;yla 2 sô khac nhau x^2+2y^2+2xy+3x+3y-4=0 tim gtnn va GTLN cua A=x^2+y^2
Tim GTNN
B=x^2+y^2+2x-y+1
C=2x^2+y^2+2xy+2x-2y+8