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

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

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

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

\(A=2x^2+y^2+2xy-6x-2y+10\)

<=>\(A=y^2+2y\left(x-1\right)+2x^2-6x+10\)

<=>\(A=y^2+2y\left(x-1\right)+\left(x^2-2x+1\right)+\left(x^2-4x+4\right)+5\)

<=>\(A=y^2+2y\left(x-1\right)+\left(x-1\right)^2+\left(x-2\right)^2+5\)

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

=> A đạt giá trị nhỏ nhất là 5 khi \(\hept{\begin{cases}\left(y+x-1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y+x-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}\)

a) \(A=4x^2-12x+100=\left(2x\right)^2-12x+3^2+91=\left(2x-3\right)^2+91\)

Ta có: \(\left(2x-3\right)^2\ge0\forall x\inℤ\)

\(\Rightarrow\left(2x-3\right)^2+91\ge91\)

hay A \(\ge91\)

Dấu "=" xảy ra <=> \(\left(2x-3\right)^2=0\)

<=> 2x-3=0

<=> 2x=3

<=> \(x=\frac{3}{2}\)

Vậy Min A=91 đạt được khi \(x=\frac{3}{2}\)

b) \(B=-x^2-x+1=-\left(x^2+x-1\right)=-\left(x^2+x+\frac{1}{4}-\frac{5}{4}\right)=-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)

Ta có: \(-\left(x+\frac{1}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\) hay B\(\le\frac{5}{4}\)

Dấu "=" \(\Leftrightarrow-\left(x+\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy Max B=\(\frac{5}{4}\)đạt được khi \(x=\frac{-1}{2}\)

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

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

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

Ta có: 

\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x;y\inℤ\\x^2\ge0\forall x\inℤ\end{cases}}\)

\(\Leftrightarrow\left(x+y-1\right)^2+x^2+1\ge1\)

hay C\(\ge\)1

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

Vậy Min C=1 đạt được khi y=1 và x=0

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

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

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

\(=\left(x+1+y\right)^2+\left(2x-1\right)^2+2003\ge2002\) 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 \(M_{min}=2002\)

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

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

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

\(A=x^2+2x\left(y+1\right)+\left(y+1\right)^2-\left(y+1\right)^2+2y^2-4y+2028\)

\(=\left(x+y+1\right)^2-y^2-2x-1+2y^2-4y+2028\)

\(=\left(x+y+1\right)^2-6x+y^2+2027\)

\(=\left(x+y+1\right)+\left(y-3\right)^2+2018\ge2018\forall x;y\) (do...)

=> MinA = 2018 \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)

Ta có :

\(x^2+y^2+2x+2y+2xy+5\)

\(=\left(x^2+2xy+y^2\right)+2\left(x+y\right)+5\)

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

Đặt x+y=a

Biểu thức trở thành :

\(a^2+2a+5\)

\(=a^2+2a+1+4\)

\(=\left(a+1\right)^2+4\)

Vì \(\left(a+1\right)^2\ge0\)

\(\Rightarrow\left(a+1\right)^2+4\ge4\)

Dấu " = " xảy ra khi a + 1 = 0

<=> x+y+1=0

Vậy biểu thức đạt giá trị nhỏ nhất là 4 khi x + y + 1 = 0

 x^2 - 2xy + 6y^2 - 12x + 2y +45 
= x^2 - 2x(y+6) + (y+6)^2 - (y+6)^2 + 6y^2 +2y + 45 
= (x - y - 6)^2 - y^2 - 12y - 36 + 6y^2 + 2y + 45 
= (x - y - 6)^2 + 5y^2 - 10y + 9 
= (x - y - 6)^2 + 5.(y^2 - 2y +1) + 4 
= (x - y - 6)^2 + 5.(y-1)^2 + 4 
=>> MIN=4 khi (x;y)={(7;1)} 

bằng 5 nhé bạn