\(A=9x^2+2y^2+6xy-6x+11\)

=> \(A=9x^2+6x\left(y-1\right)+2y^2+11\)

=> \(A=\left(3x\right)^2+2.3x\left(y-1\right)+\left(y-1\right)^2-\left(y-1\right)^2+2y^2+11\)

=> \(A=\left(3x+y-1\right)^2-\left(y^2-2y+1\right)+2y^2+11\)

=> \(A=\left(3x+y-1\right)^2-y^2+2y-1+2y^2+11\)

=> \(A=\left(3x+y-1\right)^2+y^2+2y+1+9\)

=> \(A=\left(3x+y-1\right)^2+\left(y+1\right)^2+9\)

Có \(\left(3x+y-1\right)^2\ge0\)với mọi x; y

\(\left(y+1\right)^2\ge0\)với mọi y

=> \(\left(3x+y-1\right)^2+\left(y+1\right)^2+9\ge9\)với mọi x; y

=> \(A\ge9\)với mọi x; y

Dấu "=" xảy ra <=> 3x + y - 1 = 0 và y + 1 = 0

<=> 3x + y = 1 và y = -1

<=> x = -4 và y = -1

KL: A_min = 9 <=> x = -4 và y = -1

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

\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)

\(Q_{max}=\dfrac{25}{2}\) khi \(x=\dfrac{3}{2}\)

\(A=\dfrac{9\left(x^2+2\right)-9x^2+6x-1}{x^2+2}=9-\dfrac{\left(3x-1\right)^2}{x^2+2}\le9\)

\(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

\(A=\dfrac{12x+34}{2\left(x^2+2\right)}=\dfrac{-\left(x^2+2\right)+x^2+12x+36}{2\left(x^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(x+6\right)^2}{2\left(x^2+2\right)}\le-\dfrac{1}{2}\)

\(A_{min}=-\dfrac{1}{2}\) khi \(x=-6\)

Bài 1: 

a) Ta có: \(A=-x^2-4x-2\)

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

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

\(=-\left(x+2\right)^2+2\le2\forall x\)

Dấu '=' xảy ra khi x=-2

b) Ta có: \(B=-2x^2-3x+5\)

\(=-2\left(x^2+\dfrac{3}{2}x-\dfrac{5}{2}\right)\)

\(=-2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}\right)\)

\(=-2\left(x+\dfrac{3}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{4}\)

c) Ta có: \(C=\left(2-x\right)\left(x+4\right)\)

\(=2x+8-x^2-4x\)

\(=-x^2-2x+8\)

\(=-\left(x^2+2x-8\right)\)

\(=-\left(x^2+2x+1-9\right)\)

\(=-\left(x+1\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=-1

Bài 2: 
a) Ta có: \(=25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3>0\forall x\)

b) Ta có: \(B=9x^2-6xy+2y^2+1\)

\(=9x^2-6xy+y^2+y^2+1\)

\(=\left(3x-y\right)^2+y^2+1>0\forall x,y\)

c) Ta có: \(E=x^2-2x+y^2-4y+6\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+1>0\forall x,y\)