Đề đúng: \(C=x^2+4y^2+2x-4y-4xy+2011\)

\(C=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1+2010\)

\(C=\left(x-2y\right)^2+2\left(x-2y\right)+1+2010\)

\(C=\left(x-2y+1\right)^2+2010\ge2010\)

Dấu "=" xảy ra khi: \(\left(x-2y+1\right)^2=0\)

=> Ta có vô số cặp (x;y) thỏa mãn ví dụ như:

(1;1) ; (-1;0) ; (3;2) ; ...

C = x² + 4y² + 2x - 4y - 4xy + 2011 ( đúng chưa :v )

C = [ ( x² - 4xy + 4y² ) + 2x - 4y + 1 ] + 2010

C = [ ( x - 2y )² + 2( x - 2y ) + 1 ] + 2010

C = [ ( x - 2y ) + 1 ]² + 2010

C = ( x - 2y + 1 )² + 2010 ≥ 2010 ∀ x,y 

Đẳng thức xảy ra <=> x - 2y + 1 = 0

                            <=> x - 2y = -1

                            <=> x = 2y - 1

=> MinC = 2011 <=> x = 2y - 1

\(A=2x^2+4y^2+4xy+2x+4y+9\)

\(=2\left(x^2+x\left(2y+1\right)+\dfrac{\left(2y+1\right)^2}{4}\right)-\dfrac{\left(2y+1\right)^2}{2}+4y^2+4y+9\)

\(=2\left(x+\dfrac{2y+1}{2}\right)^2-2y^2-2y-\dfrac{1}{2}+4y^2+4y+9\)

\(=2\left(x+\dfrac{2y+1}{2}\right)^2+2y^2+2y+\dfrac{17}{2}\)

\(=2\left(x+\dfrac{2y+1}{2}\right)^2+2\left(y+\dfrac{1}{2}\right)^2+8\ge8\)

Dấu '' = '' xảy ra khi: \(\Leftrightarrow\left\{{}\begin{matrix}y+\dfrac{1}{2}=0\\x+\dfrac{2y+1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=0\end{matrix}\right.\)

Vậy: Min A = 8 khi \(x=0;y=-\dfrac{1}{2}\)

yiouoiyy

\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)

Vì \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-5\\y=-3\\z=8\end{cases}}}\)

\(A=2x^2+4y^2+4xy+2x+4y+9=\left(x^2+4y^2+4xy+2x+4y+1\right)+x^2+8\)

   \(=\left(x+2y+1\right)^2+x^2+8\ge8\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+2y+1=0\\x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=-\frac{1}{2}\end{cases}}}\)

Vậy \(Min\left(A\right)=8\Leftrightarrow\hept{\begin{cases}x=0\\y=-\frac{1}{2}\end{cases}}\)

P = 2x^x+4y²+4xy+2x+4y+9

   = x²+(x²+4y²+1+4xy+2x+4y) +8

   = x²+(x+2y+1)²+8 \(\ge\)8

dấu bằng xảy ra khi x=0 y=-0.5

\(A=x^2+5y^2-4xy-2x-4y+5=x^2-2x\left(2y+1\right)+\left(2y+1\right)^2+\left(y^2-8y+16\right)-12=\left(x-2y-1\right)^2+\left(y-4\right)^2-12\ge-12\)

\(minA=-12\Leftrightarrow\)\(\left\{{}\begin{matrix}x=9\\y=4\end{matrix}\right.\)

\(A=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(x^2-4x+4\right)-3\)

\(A=\left(x-2y+1\right)^2+\left(x-2\right)^2-3\ge-3\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;\dfrac{3}{2}\right)\)