1/ x^2 +4xy +4y^2 = (x +2y)^2

2/ -x^3 +9x^2 -27x+27= - (x^3 -9x^2+27x-27) = - (x-3)^3

3/ 8x^6 +36x^4y+54^2y^2+27y^3 = (2x^2+3y)^3

4/ x^3 - 6x^2y+12xy^2 -8y^3= (x-2y)^3

1) x² + 4xy + 4y² = ( x + 2y )²

2) - x³ + 9x² - 27x + 27 = ( 3 - x )²

3) 8x⁶ + 36x⁴y + 54x²y² + 27y³ = ( 2x² + 3y )³

4) x³ - 6x²y + 12xy² - 8y³ = ( x - 2y )³

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

6) x² + y² + 4 + 2xy + 4x + 4y = ( x + y + 2 )²

a: Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)

b: Ta có: \(x^2+y^2-4x+y+5\)

\(=\left(x^2-4x+4\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi x=2 và \(y=-\dfrac{1}{2}\)

1) \(4x^5y^2-8x^4y^2+4x^3y^2\)

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

\(=4x^3y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

\(=4x^3y^2\left(x-1\right)^2\)

2) \(5x^4y^2-10x^3y^2+5x^2y^2\)

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

\(=5x^2y^2\left(x^2-2\cdot x\cdot1+1^2\right)\)

\(=5x^2y^2\left(x-1\right)^2\)

3) \(12x^2-12xy+3y^2\)

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

\(=3\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

4) \(8x^3-8x^2y+2xy^2\)

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

\(=2x\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

5) \(20x^4y^2-20x^3y^3+5x^2y^4\)

\(=5x^2y^2\left(4x^2-4xy+y^2\right)\)

\(=5x^2y^2\left[\left(2x\right)^2-2\cdot2x\cdot y+y^2\right]\)

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

1: 4x^5y^2-8x^4y^2+4x^3y^2

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

=4x^3y^2(x-1)^2

2: \(=5x^2y^2\left(x^2-2x+1\right)=5x^2y^2\left(x-1\right)^2\)

3: \(=3\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)^2\)

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

5: \(=5x^2y^2\left(4x^2-4xy+y^2\right)=5x^2y^2\left(2x-y\right)^2\)

3, \(C=x^2-8xy+16y^2\)

\(C=x^2-2\cdot4y\cdot x+\left(4y\right)^2\)

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

Thay \(x-4y=5\) vào C ta được:

\(C=5^2=25\)

Vậy: ......

4, \(D=9x^2+1620-12xy+4y^2\)

\(D=\left(9x^2-12xy+4y^2\right)+1620\)

\(D=\left[\left(3x\right)^2-2\cdot3x\cdot2y+\left(2y\right)^2\right]+1620\)

\(D=\left(3x-2y\right)^2+1620\)

Thay \(3x-2y=20\) vào D ta được:

\(D=20^2+1620=400+1620=2020\)

Vậy: ...

3/

\(C=x^2-8xy+16y^2=x^2-2.4.xy+\left(4y\right)^2=\left(x-4y\right)^2\)

Thay x - 4y =  5 ta có: \(C=5^2=25\)

4/

\(D=9x^2-12xy+4y^2+1620\\ =\left(3x\right)^2-3.2.2xy+\left(2y\right)^2+1620\\ =\left(3x-2y\right)^2+1620\)

Thay 3x - 2y = 20. Ta có: \(D=20^2+1620=400+1620=2020\)

a, bậc 6 

b, bậc 6 

c, bậc 12 

d, bậc 9 

e, bậc 8