CMR B \(\ge\)một số nào đó

\(B=2x^2+2xy+y^2-2x+2y+2016\)

\(=\left(x^2+2xy+y^2+2x+2y+1\right)+\left(x^2-4x+4\right)+2011\)

\(=\left[\left(x+y\right)^2+2\left(x+y\right)+1\right]+\left(x-2\right)^2+2011\)

\(=\left(x+y+1\right)^2+\left(x-2\right)^2+2011\ge2011\forall x;y\)có GTNN là 2011

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy \(B_{min}=2011\) tại \(x=2;y=-3\)

thang ngu

\(D=\left(x^2+z^2-2xz\right)+\left(x^2+y^2-2xy+2x-2y+1\right)+3\)

\(D=\left(x-z\right)^2+\left(x-y+1\right)^2+3\ge3\)

\(D_{min}=3\) khi \(\left\{{}\begin{matrix}x=z\\x=y-1\end{matrix}\right.\)

C = 2x² + 2xy  + y² - 2x -2y +2 

= x² + 2x(y - 1) + (y - 1)² + x² + 1 

= (x + y - 1)² + x² + 1

Tới đây tự làm

Ta có:

A = 2x² + 2xy + y² - 2x + 2y + 2

A = (x² + 2xy + y²) + 2(x + y) + 1 + (x² - 4x + 4) - 3

A = (x + y)² + 2(x + y) + 1 + (x - 2)² - 3

A = (x + y + 1)² + (x - 2)² - 3 \(\ge\)-3 \(\forall\)x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y+1=0\\x-2=0\end{cases}}\) <=> \(\hept{\begin{cases}y=-x-1\\x=2\end{cases}}\) <=> \(\hept{\begin{cases}y=-2-1=-3\\x=2\end{cases}}\)

Vậy MinA = -3 <=> x = 2 và y = -3

\(2x^2+2xy+y^2-2x+2y+\)\(2\)

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

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

Ta thấy \(\left(x+y+1\right)^2\ge0\)     \(\forall x,y\)

             \(\left(x-2\right)^2\ge0\)            \(\forall x\)

=> \(\left(x+y+1\right)^2+\left(x-2\right)^2\ge0\)     \(\forall x,y\)

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

hay \(A\ge-1\)

\(MinA=-1\)\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

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

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

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

Dấu \(=\)khi \(\hept{\begin{cases}x-2=0\\x+y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}\).

x^2+y^2+2x+2y+2xy+5

bài này rễ quá

giá trị nhỏ nhất của A là 5 nhé

\(S=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(y^2-4y+4\right)+2021\)

\(S=\left(x+y+1\right)^2+\left(y-2\right)^2+2021\ge2021\)

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

\(M=5x^2+y^2-2x+2y+2xy+2004\)

\(=\left(x^2+2x+1\right)+2y\left(x+1\right)+y^2+4x^2-4x+1+2002\)

\(=\left(x+1\right)^2+2y\left(x+1\right)+y^2+\left(2x-1\right)^2+2002\)

\(=\left(x+1+y\right)^2+\left(2x-1\right)^2+2003\ge2002\) với mọi x,y

=> \(M_{min}=2002\Leftrightarrow\left\{{}\begin{matrix}x+y+1=0\\2x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(M_{min}=2002\)