=(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

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

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

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

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

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

\(a,4x^2+9y^2+4x-24y+17=0\)

\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)

\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)

\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)

  giúp mình với chiều thì rồi

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)\(B=x^2-2xy+2y^2-x+y\)????

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

xem lại đề

\(D=x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy GTNN của D là \(\frac{3}{4}\)khi x = \(\frac{1}{2}\)

\(E=x\left(x-3\right)=x^2-3x=\left(x^2-3x+\frac{9}{4}\right)-\frac{9}{4}=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)

Vậy GTNN của E là \(-\frac{9}{4}\)khi x = \(\frac{3}{2}\)

\(G=x^2+5y^2+2xy-2y+100\)

\(G=\left(x^2+2xy+y^2\right)+\left(4y^2-2y+\frac{1}{4}\right)+\frac{399}{4}\)

\(G=\left(x+y\right)^2+\left(2y-\frac{1}{2}\right)^2+\frac{399}{4}\ge\frac{399}{4}\)

Vậy GTNN của G là \(\frac{399}{4}\)khi x = \(-\frac{1}{4}\); y = \(\frac{1}{4}\)

Hihiii cam on bann nhieuu nhe <3