a) \(x^3y^3+x^2y^2+4\)

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

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

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

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

b) \(x^3+3x^2y-9xy^2+5y^3\)

\(=x^3+5x^2y-2x^2y-10xy^2+xy^2+5y^3\)

\(=\left(5y^3-10xy^2+5x^2y\right)+\left(xy^2-2x^2y+x^3\right)\)

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

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

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

x³ + 3x²y - 9xy² + 5y³

= ( x³ - 3x²y + 3xy² - y³ ) + ( 6y³ - 12xy² + 6 x²y )

= ( x - y )³ + 6y ( x - y )²

= ( x - y )² ( x + 5y )

Chon D

D

1: \(6x^2y-9xy^2+3xy\)

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

2: \(\left(4-x\right)^2-16\)

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

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

\(=x^2\left(x+9\right)-4\left(x+9\right)\)

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

1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)

2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

phân tích đa thức thành nhân tử

a, 6x^2 + 7xy + 2y^2

=6x^2+3xy+4xy+2y^2

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

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

b, 9x^2 - 9xy - 4y^2

=9x^2 +3xy-12xy-4y^2

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

=(3x+4y)(x+y)

c, x^2 - y^2 + 10x - 6y + 16=x^2-y^2+6x-6y+4x+16=x(x+6)-y(x+6)+4(x+6)=(x-y+4)(x+6)

Bài làm

a, 6x² + 7xy + 2y²

= 6x² + 3xy + 4xy + 2y² 

= ( 6x² + 3xy ) + ( 4xy + 2y² )

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

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

b, 9x² - 9xy - 4y² 

= 9x² - 12xy + 3xy - 4y² 

= ( 9x² - 12xy ) + ( 3xy - 4y² )

= 3x( 3x - 4y ) + y ( 3x - 4y )

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

c, x² - y² + 10x - 6y + 16

 = x² - y² - 6x + 6y + 4x + 16

= x( x + 6 ) - y( x + 6 ) + 4( x + 6 )

= ( x - y + 4 )( x + 6 )

# Học tốt #

\(\Rightarrow\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\cdot\left(x^2+2x\right)+20=\left(x^2+2x\right)\left(x^2+2x+4\right)+5\left(x^2+2x+4\right)=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)

Lời giải:

a.

$=\frac{1}{2}(x^2-4y^2)=\frac{1}{2}[x^2-(2y)^2]=\frac{1}{2}(x-2y)(x+2y)$

b.

$=\frac{1}{3}x(y+3xz+3z)$

c.

$=\frac{2}{25}x(225x^2-4)=\frac{2}{25}(15x-2)(15x+2)$

d.

$=\frac{1}{5}x^2(2+25x+5y)$

\(\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\left(x^2+2x\right)+20\)

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

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

(x2+2x)+9x2+18x+20

=(x2+2x)+9(x2+2x)+20

Đặt t=x2+2x đc:

t+9t+20=10t+20=10(t+2)

Thay t=x2+2x vào đc:

10(x2+2x+2)