Ta có: 

\(C=\sqrt{-x^2+6x}\) 

Mà: \(\sqrt{-x^2+6x}\ge0\) 

Dấu "=" xảy ra khi:

\(\sqrt{-x^2+6x}=0\)

\(\Leftrightarrow\sqrt{-x\left(x-6\right)}=0\)

\(\Leftrightarrow-x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

Vậy: \(C_{min}=0\) khi \(\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

\(D=\sqrt{6x-2x^2}\)

Mà: \(\sqrt{6x-2x^2}\ge0\)

Dấu "=" xảy ra khi:

\(\sqrt{6x-2x^2}=0\)

\(\Leftrightarrow\sqrt{2x\left(3-x\right)}=0\)

\(\Leftrightarrow2x\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy: \(D_{min}=0\) khi \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(C=\sqrt{-x^2+6x}=\sqrt{9-\left(x^2-6x+9\right)}=\sqrt{9-\left(x-3\right)^2}\le\sqrt{9}=3\)

Dấu "=" xảy ra khi \(x=3\)

Vậy \(maxC=3\)

\(D=\sqrt{6x-2x^2}=\dfrac{1}{\sqrt{2}}\sqrt{12x-4x^2}=\dfrac{1}{\sqrt{2}}\sqrt{9-\left(4x^2-12x+9\right)}\)

\(=\dfrac{1}{\sqrt{2}}\sqrt{9-\left(2x-3\right)^2}\le\dfrac{1}{\sqrt{2}}.\sqrt{9}\)\(=\dfrac{3\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(x=\dfrac{3}{2}\)

Vậy \(maxD=\dfrac{3\sqrt{2}}{2}\)

Lời giải:

$A=(9x^2+6xy+y^2)+y^2-6x+4y+17$

$=(3x+y)^2-2(3x+y)+y^2+6y+17$

$=(3x+y)^2-2(3x+y)+1+(y^2+6y+9)+7$

$=(3x+y-1)^2+(y+3)^2+7\geq 0+0+7=7$

Vậy GTNN của biểu thức là $7$. Giá trị này đạt được khi $3x+y-1=y+3=0$

$\Leftrightarrow y=-3; x=\frac{4}{3}$

$A$ không có max bạn nhé.

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

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

Vậy Min_P=2 khi x=1, y=-3

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

`A=(2x)^2+2.2x.1+1^2+1=(2x+1)^2+1`

`=> A_(min)=1 <=>x=-1/2`

`B=(\sqrt2x)^2-2.\sqrt2 x . \sqrt2/2 + (\sqrt2/2)^2 + 1/2`

`=(\sqrt2x-\sqrt2/2)^2+1/2`

`=> B_(min)=1/2 <=> x=1/2`

`C=-(x^2-2.x.3+3^2+6)=-(x-3)^2-6`

`=> C_(max)=-6 <=> x=3`