\(M=2x^2+9y^2-6xy-6x-12y+2028\\ =3\left(x^2-2xy+y^2\right)-\left(x^2+6x+9\right)+6\left(y^2-2y+1\right)+2025\\ =\left(x-y\right)^2-\left(x-3\right)^2+6\left(y-1\right)^2+2025\ge2025\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=1\end{matrix}\right.\) (vô lí) nên dấu \("="\) ko thể xảy ra

\(N=x^2-4xy+5y^2+10x-22y+28\\ =\left(x^2+4y^2+25-4xy-20y+10x\right)+\left(y^2-2y+1\right)+2\\=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-2y=5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

\(M=2x^2+9y^2-6xy-6x-12y+2028=\left(x+2\right)^2-6y\left(x+2\right)+9y^2+\left(x-5\right)^2+1999=\left(x+2-3y\right)^2+\left(x-5\right)^2+2019\ge1999\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\dfrac{7}{3}\end{matrix}\right.\)

\(N=x^2-4xy+5y^2+10x-22y+28=\left(x+5\right)^2-4y\left(x+5\right)+4y^2+\left(y-1\right)^2+2=\left(x+5-2y\right)^2+\left(y-1\right)^2+2\ge2\)

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

\(2x^2+9y^2-6xy-6x-12y+2004\)

\(=x^2-10x+25+x^2+9y^2+4-6xy+4x-12y+1975\)

\(=\left(x-5\right)^2+\left(x-3y+2\right)^2+1975\ge1975\)

Dấu \(=\)khi \(\hept{\begin{cases}x-5=0\\x-3y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\).

A = -x² + 2xy - 4y² + 2x + 10y - 8

=> -A = x² - 2xy + 4y² - 2x - 10y + 8

          = ( x² - 2xy + y² - 2x + 2y + 1 ) + ( 3y² - 12y + 12 ) - 5

          = [ ( x² - 2xy + y² ) - ( 2x - 2y ) + 1 ] + 3( y² - 4y + 4 ) - 5

          = [ ( x - y )² - 2( x - y ) + 1 ] + 3( y - 2 )² - 5

          = ( x - y - 1 )² + 3( y - 2 )² - 5 ≥ -5 ∀ x, y

Dấu "=" xảy ra <=> x = 3 ; y = 2

=> -A ≥ -5

=> A ≤ 5

=> MaxA = 5 <=> x = 3 ; y = 2

B = 2x² + 9y² - 6xy - 6x - 12y + 2004

= ( x² - 6xy + 9y² + 4x - 12y + 4 ) + ( x² - 10x + 25 ) + 1975

= [ ( x² - 6xy + 9y² ) + ( 4x - 12y ) + 4 ] + ( x - 5 )² + 1975

= [ ( x - 3y )² + 2( x - 3y ).2 + 2² ] + ( x - 5 )² + 1975

= ( x - 3y + 2 )² + ( x - 5 )² + 1975 ≥ 1975 ∀ x, y

Dấu "=" xảy ra <=> x = 5 ; y = 7/3

=> MinB = 1975 <=> x = 5 ; y = 7/3

Ta có: A = -x² + 2xy - 4y² + 2x + 10y - 8

A = -[x² - 2xy + 4y² - 2x - 10y + 8]

A = -[(x² - 2xy + y²) - 2(x + y) + 1 + 3y² - 12y + 12 - 5]

A = -[(x - y)² - 2(x + y) + 1 + 3(y - 2)²]+ 5

A = -[(x - y - 1)² + 3(y - 2)²] + 5 \(\le\) 5 với mọi x

Dấu "=" xảy ra <=> x - y - 1 = 0 và y + 2 = 0

=>x = -1 và y = -2

Vậy MaxA = 5 khi x = -1 và y = -2

B = 2x² + 9y² - 6xy - 6x - 12y + 2004

B = (x² - 6xy + 9y²) + 4(x - 3y) + 4 + x² - 10x + 25 + 1975

B = (x - 3y + 2)² + (x - 5)² + 1975 \(\ge\)1975

đoạn cuối tt trên

Nhân thêm 2 vào F là mọi việc sẽ ez bởi hằng đẳng thức cơ bản:D