1. x² - 16 - 4xy + 4y²

= ( x² - 4xy + 4y² ) - 16

= ( x - 2y )² - 4²

= ( x - 2y - 4 )( x - 2y + 4 )

2. 4x² + 4x - 3

= ( 4x² + 4x + 1 ) - 4

= ( 2x + 1 )² - 2

= ( 2x + 1 - 2 )( 2x + 1 + 2 )

= ( 2x - 1 )( 2x + 3 )

3. x² - x - 12

= x² + 3x - 4x - 12

= x( x + 3 ) - 4( x + 3 )

= ( x + 3 )( x - 4 )

4. 3x + 3y - x² - 2xy - y²

= ( 3x + 3y ) - ( x² + 2xy + y² )

= 3( x + y ) - ( x + y )²

= ( x + y )( 3 - x - y )

5. 4y⁴ + 16 

= 4( y⁴ + 4 )

= 4( y⁴ + 4y² + 4 - 4y² )

= 4[ ( y⁴ + 4y² + 4 ) - 4y² ]

= 4[ ( y² + 2 )² - ( 2y )² ]

= 4( y² - 2y + 2 )( y² + 2y + 2 )

a,\(x^2-16-4xy+4y^2\)

\(=\left(x^2-4xy+4y^2\right)-16\)

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

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

b,\(4x^2+4x-3\)

\(=4x^2-2x+6x-3\)

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

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

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

c,\(x^2-x-12\)

\(=x^2-4x+3x-12\)

\(=\left(x^2+3x\right)-\left(4x-12\right)\)

\(=x\left(x+3\right)-4\left(x+3\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

2x - 2x² - 5

= -2( x² - x + 1/4 ) - 9/2

= -2( x - 1/2 )² - 9/2 ≤ -9/2 ∀ x

Dấu "=" xảy ra <=> x = 1/2

Vậy GTLN của biểu thức = -9/2 <=> x = 1/2

x-x\(^2\)=0

Ta có:

\(x-x^2\)

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

\(=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(-\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)

Vậy Max = 1/4 khi x = 1/2

Ta có: \(4x-x^2+3\)

\(=-\left(x^2-4x+4\right)+7\)

\(=-\left(x-2\right)^2+7\le7\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(-\left(x-2\right)^2=0\Rightarrow x=2\)

Vậy Max = 7 khi x = 2

4x - x² + 3

= -( x² - 4x + 4 ) + 7

= -( x - 2 )² + 7 ≤ 7 ∀ x

Dấu = xảy ra <=> x = 2

Vậy GTLN của đa thức = 7 <=> x = 2

Ta có: 

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

\(M=\left(x^2-x+\frac{1}{4}\right)+\left(y^2+6y+9\right)+\frac{3}{4}\)

\(M=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-3\end{cases}}\)

M = x² + y² - x + 6y + 10

= ( x² - x + 1/4 ) + ( y² + 6y + 9 ) + 3/4

= ( x - 1/2 )² + ( y + 3 )² + 3/4 ≥ 3/4 ∀ x

Dấu "=" xảy ra <=> x = 1/2 ; y = -3

=> MinM = 3/4 <=> x = 1/2 ; y = -3

Ta có 5x² - (2x + 1).(x - 2) - x(3x + 3) + 7

= 5x² - (2x² - 3x - 2) - 3x² - 3x + 7

= 5x² - 2x² + 3x + 2 - 3x² - 3x + 7

= 9 \(\forall\)x

=> Biểu thức không phụ thuộc vào biến 

5x² - ( 2x + 1 )( x - 2 ) - x( 3x + 3 ) + 7

= 5x² - ( 2x² - 3x - 2 ) - 3x² - 3x + 7

= 5x² - 2x² + 3x + 2 - 3x² - 3x + 7

= 9

=> đpcm

x = 0

dung KO 

thi ket ban voi minh nha OK

Ta có: \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)

Thay \(x+y=2\)và \(xy=-2\)vào biểu thức ta được:

\(x^3+y^3=2^3-3.\left(-2\right).2=20\)

x³ + y³ = x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy²

            = ( x³ + 3x²y + 3xy² + y³ ) - ( 3x²y + 3xy² )

            = ( x + y )³ - 3xy( x + y )

            = 2³ - 3.(-2).2

            = 8 + 12 = 20

D = x² - 2x + y² - 4y + 6

= ( x² - 2x + 1 ) + ( y² - 4y + 4 ) + 1

= ( x - 1 )² + ( y - 2 )² + 1

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu "=" xảy ra <=> x = 1 ; y = 2

=> MinD = 1 <=> x = 1 ; y = 2

\(D=x^2-2x+y^2-4y+6\)

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

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

Vì \(\left(x-1\right)^2\ge0\forall x\); \(\left(y-2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

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

Vậy \(A_{min}=1\)\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)