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biet tong cua so thu nhat va so thu hai bang 5,8.Tong cua so thu hai va so thu ba bang 6,7.Tong so thu nhat va so thu ba bang 7,5.Tim moi so do?
\(A=x^2+2xy+2y^2+2x-4y+2013\)
\(=\left(x^2+y^2+1+2x+2y+2xy\right)-1-2y+y^2-4y+2013\)\(=\left(x+y+1\right)^2+\left(y^2-2.y.3+9\right)-9+2012\)
\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2003\)
mà \(\left(x+y+1\right)^2,\left(y-3\right)^2\ge0\)
\(\Rightarrow A=x^2+2xy+2y^2+2x-4y+2013=\left(x+y+1\right)^2+\left(y-3\right)^2+2003\ge2003\)
\(\Rightarrow Min\left(A\right)=2003\)
\(A=2x^2+y^2-2xy-2x+3\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+2\)
\(A=\left(x-y\right)^2+\left(x-1\right)^2+2\)
Mà \(\left(x-y\right)^2\ge0\forall x;y\)
\(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow A\ge2\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x-y=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=1\end{cases}}\)
Vậy Min A = 2 khi x=y=1
a, B=x2+4xy+y2+x2-8x+16+2012
B=(x+y) 2+(x-4)2+2012
Vậy B >=2012 ( Dấu "=" xảy ra khi x=4,y=-4)
b làm tương tự
c, 9x2+6x+1+y2-4y+4+x2-4xz+4z2=0
(3x+1)2+(y-4)2+(x-2z)2=0
Vậy 3x+1=0 => x = -1/3
y-4=0 => y=4
x-2z=0 thế x=-1/3 ta được. -1/3-2z=0 => z = -1/6
Bạn nhớ ghi lại đề minh không ghi đề
a) \(B=2x^2+y^2+2xy-8x+2028\)
\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+4^2\right)+2012=\left(x+y\right)^2+\left(x-4\right)^2+2012\ge2012\)
\(MinB=2012\Leftrightarrow\hept{\begin{cases}x=4\\y=-4\end{cases}}\)
b)\(C=x^2+5y^2+4xy+2x+2y-7\)
\(=\left(x^2+4xy+4y^2\right)+\left(2x+4y\right)+1+\left(y^2-2y+1\right)-9\)
\(=\left(\left(x+2y\right)^2+2\left(x+2y\right)+1\right)+\left(y-1\right)^2-9=\left(x+2y+1\right)^2+\left(y-1\right)^2-9\ge9\)
\(MinC=-9\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)
c)\(10x^2+y^2+4z^2+6x-4y-4xz+5=0\)
\(\Leftrightarrow\left(9x^2+6x+1\right)+\left(y^2-4y+4\right)+\left(x^2-4xz+4z^2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)^2+\left(y-2\right)^2+\left(x-2z\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}3x+1=0\\y-2=0\\x-2z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{3}\\y=2\\z=-\frac{1}{6}\end{cases}}\)
\(A=\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(y^2-6y+9\right)+2006\)\(=\left(x+y\right)^2+2\left(x+y\right)+1+\left(y-3\right)^2+2006\)
\(=\left(x+y+1\right)^2+\left(y-3\right)^2+2006\)
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 \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)
\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
A= x2+2y2+2xy+2x-4y+2018
= x2+y2+1+2xy+2x+2y + y2-6y+9 +2008
= (x2+y2+12+2xy+2x+2y)+(y2-6y+9)+2008
= (x+y+1)2+(y-3)2+2008
Vậy GTNN của A là 2008
cứ làm bình tĩnh không lên ôm đồm
\(A=x^2+2y^2+2xy+2x-4y+2018\)
\(A_1=\left(x^2+y^2+2xy\right)+\left(2x+2y\right)+y^2-6y+2018\)
\(A_2=\left(x+y\right)^2+2\left(x+y\right)+1+\left(y^2-6y+9\right)+2018-9-1\)
\(A_4=\left(x+y+1\right)^2+\left(y-3\right)^2+2018-10\)
\(\left\{{}\begin{matrix}\left(x+y+1\right)^2\ge0\\\left(y-3\right)^2\ge0\\A\ge2008\end{matrix}\right.\)