=(x^2+y^2+2xy​)+(2x+2y)+3

=((x+y)² +2(x+y) +1)+2

=(x+y+1)²+2

vậy A_min=2

\(A=x^2+y^2+2xy+2x+2y+3\)

<=>\(A=x^2+2x\left(y+1\right)+y^2+2y+3\)

<=>\(A=x^2+2x\left(y+1\right)+\left(y^2+2y+1\right)+2\)

<=>\(A=x^2+2x\left(y+1\right)+\left(y+1\right)^2+2\)

<=>\(A=\left(x+y+1\right)^2+2\ge2\)

6 nha bạn

\(A=x^2-2xy+2y^2+2x-10y+2033\\ =x^2-2xy+y^2+y^2+2x-8y-2y+1+16+2016\\ =\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+1+\left(y^2-8y+16\right)+2016\\ =\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-4\right)^2+2016\\ =\left[\left(x-y\right)^2+2\left(x-y\right)+1\right]+\left(y-4\right)^2+2016\\ =\left(x-y+1\right)^2+\left(y-4\right)^2+2016\\ Do\text{ }\left(y-4\right)^2\ge0\forall y\\ \left(x-y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-y+1\right)^2+\left(y-4\right)^2\ge0\forall x;y\\ \Rightarrow A=\left(x-y+1\right)^2+\left(y-4\right)^2+2016\ge2016\forall x;y\\ Dấu\text{ }''=''\text{ }xảy\text{ }ra\text{ }khi:\left\{{}\begin{matrix}\left(y-4\right)^2=0\\\left(x-y+1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y-4=0\\x-y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x-4+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\\ Vậy\text{ }A_{\left(Min\right)}=2016\text{ }khi\text{ }\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

xem lại đề

a/ \(2x^2+8x+1=2\left(x^2+4x+\frac{1}{2}\right)=2\left(x^2+2.2x+4-4+\frac{1}{2}\right)\)

     \(=2\left[\left(x+2\right)^2-\frac{7}{2}\right]=2\left(x+2\right)^2-7\ge-7\)

       Vậy Min A = -7 khi x + 2 = 0 => x = 2

b/ \(2x^2+3x+1=2\left(x^2+\frac{3}{2}x+\frac{1}{2}\right)=2\left(x^2+2.\frac{3}{4}.x+\frac{9}{16}-\frac{9}{16}+\frac{1}{2}\right)\)

          \(=2\left[\left(x+\frac{3}{4}\right)^2-\frac{1}{16}\right]=2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)

         Vậy Min B = -1/8 khi x + 3/4 = 0 => x = -3/4