a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

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

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

a) (2xy+5)(4x^2+5): = 2xy * 4x^2 + 2xy * 5 + 5 * 4x^2 + 5 * 5 = 8x^3y + 10xy + 20x^2 + 25 b) (6xy+4)(2x^2+1): = 6xy * 2x^2 + 6xy * 1 + 4 * 2x^2 + 4 * 1 = 12x^3y + 6xy + 8x^2 + 4 c) (9x^2+4)(3x+5): = 9x^2 * 3x + 9x^2 * 5 + 4 * 3x + 4 * 5 = 27x^3 + 45x^2 + 12x + 20 d) (-2xy+6)(1/2xy+7): = -2xy * 1/2xy + (-2xy) * 7 + 6 * 1/2xy + 6 * 7 = -xy + (-14xy) + 3 + 42 = -15xy + 45 e) (4x+1)(2x^2+5x+2): = 4x * 2x^2 + 4x * 5x + 4x * 2 + 1 * 2x^2 + 1 * 5x + 1 * 2 = 8x^3 + 20x^2 + 8x + 2x^2 + 5x + 2 = 8x^3 + 22x^2 + 13x + 2 f) (2x^2y+3x)(2x+1): = 2x^2y * 2x + 2x^2y * 1 + 3x * 2x + 3x * 1 = 4x^3y + 2x^2y + 6x^2 + 3x g) (4xy+5x^2y)(2xy+6): = 4xy * 2xy + 4xy * 6 + 5x^2y * 2xy + 5x^2y * 6 = 8x^2y^2 + 24xy + 10x^3y + 30x^2y = 8x^2y^2 + 30x^2y + 24xy h) (-1/2x^2+6)(4xy+5): = -1/2x^2 * 4xy + (-1/2x^2) * 5 + 6 * 4xy + 6 * 5 = -2xy + (-5/2x^2) + 24xy + 30 = 22xy + (-5/2x^2) + 30

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)

D= 5x^2+8xy+5y^2-2x+2y  

=4x^2+8xy+4y^2-2x+2y+y^2+x^2

=(2x+2y)^2+x^2-2*1/2x+1/4+y^2+2*1/2y+1/4-1/2

(2x+2y)^2+(x-1/2)^2+(y+1/2)^2-1/2>=-1/2

suy ra D>=-1/2 nên D có GTNN là -1/2

Ta có : 5D = 25x² + 40xy + 25y² - 10x + 10y

5D = (5x+ 4y - 1)² + 9y² + 18y - 1  

5D = ( 5x + 4y - 1)² + 9 (y + 1)² - 2

D =\(\frac{1}{5}\). ( 5x + 4y - 1)² + \(\frac{9}{5}\).( y + 1)²  -  \(\frac{2}{5}\)  \(\ge\)\(\frac{-2}{5}\)

Dấu "=" xảy ra khi y+1 = 0  \(\Leftrightarrow\)y = -1

                          5x + 4y - 1 = 0  \(\Leftrightarrow\)x=1

Vậy GTNN của D = \(\frac{-2}{5}\)khi x = 1 ; y = -1

Ta có: 2E = 4x² + 8y² - 8xy - 8x - 8y + 4006

2E = ( 4y - x - 1)² + x² - 6x + 4003

2E = ( 4y - x - 1)² + ( x - 3)² + 4003 - 9

E = \(\frac{1}{2}\).( 4y - x - 1)² +\(\frac{1}{2}\).( x - 3 )² + 1997 \(\ge\)1997

Dấu "=" xảy ra khi x - 3 = 0  \(\Leftrightarrow\)x = 3

                            4x - x -1 = 0  \(\Leftrightarrow\)y = 1

Vậy GTNN của E = 1997 khi x = 3 ; y = 1

a) \(3x^2-3xy-5x+5y\)

\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)

\(=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(3x-5\right)\)

b) \(2x^3y-2xy^3-4xy^2-2xy\)

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

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

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

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

c) \(x^2+1+2x-y^2\)

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

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

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

d) \(x^2+4x-2xy-4y+y^2\)

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

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

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

e) \(x^3-2x^2+x\)

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

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

f) \(2x^2+4x+2-2y^2\)

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

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

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

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

a: =3x(x-y)-5(x-y)

=(x-y)(3x-5)

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

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

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

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

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