D ez nhất :v

\(D=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+5\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+5\ge5\)

Đẳng thức xảy ra khi x = 1 và y = -2

\(A=\left[\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4\right]+\left(y^2-2y+1\right)+2020\)

\(=\left[\left(x-y\right)^2+2\left(x-y\right).2+2^2\right]+\left(y-1\right)^2+2020\)

\(=\left(x-y+2\right)^2+\left(y-1\right)^2+2020\ge2020\)

Dấu "=" xảy ra khi y = 1 và x - y + 2 = 0 tức là x = y - 2 = -1

Mình sửa lại đề nha:    \(A=2x^2+2y^2+2xy-2x+2y+2\)

Ta có:  \(A=2x^2+2y^2+2xy-2x+2y+2\)

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

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

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

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

\(\Leftrightarrow2A=4x^2+4y^2-4xy-4x+4y+4\)

\(=\left(4x^2-4xy+y^2\right)-2\left(2x-y\right)+1+3y^2+6y+4\)

\(=\left(\left(2x-y\right)^2-2\left(2x-y\right)+1\right)+\left(3y^2+6y+3\right)+2\)

\(=\left(2x-y-1\right)^2+3\left(y+1\right)^2+2\ge2\)

\(\Rightarrow A\ge2\)

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

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

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

A= (x²-2xy +y²)+(2x-2y)+1+(y²-8y+16)

A= (x-y)² +2(x-y) +1 +(y-4)²

A= (x-y+1)² +(y-4)²

Vì (x-y+1)² +(y-4)² >= 0 với mọi x,y

Dấu = xảy ra <=> x-y+1=0 và y-4=0

                   <=> x=3 và y=4

\(A=x^2+2y^2-2xy-2y-2x+2019\)

\(A=x^2+y^2+y^2-2xy+2y-4y-2x+2019\)

\(A=\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1+y^2-4y+4+2014\)

\(A=\left(x-y\right)^2-2\left(x-y\right)+1+\left(y-2\right)^2+2014\)

\(A=\left(x-y-1\right)^2+\left(y-2\right)^2+2014\ge2014\forall x;y\)

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

Bài 2: 

a: \(=-\left(x^2+2x-100\right)\)

\(=-\left(x^2+2x+1-101\right)\)

\(=-\left(x+1\right)^2+101< =101\)

Dấu = xảy ra khi x=-1

b: \(=-3\left(x^2-\dfrac{1}{3}x\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{1}{36}\right)\)

\(=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}< =\dfrac{1}{12}\)

Dấu = xảy ra khi x=1/6

c: \(=-\left(3x^2+4y^2-18x+8y-12\right)\)

\(=-\left(3x^2-18x+27+4y^2+8y+4-43\right)\)

\(=-3\left(x-3\right)^2-4\left(y+1\right)^2+43< =43\)

Dấu = xảy ra khi x=3 và y=-1