a: =(a^2-b^2)-(2a-2b)

=(a-b)(a+b)-2(a-b)

=(a-b)(a+b-2)

b: =(3x-3y)+5y(x-y)

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

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

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

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

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

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

=(-x-4y+5)(3x+2y+3)

e: =16-(x^2-4xy+4y^2)

=16-(x-2y)^2

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

g: =9x^2-6x+1-(3xy-y)

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

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

h: =(x-y)^3-z^3

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

=(x-y-z)(x^2-2xy+y^2+xz-yz+z^2)

a) \(a^2-b^2-2a+2b\)

\(=\left(a^2-b^2\right)-\left(2a-2b\right)\)

\(=\left(a+b\right)\left(a-b\right)-2\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b-2\right)\)

b) \(3x-3y-5x\left(y-x\right)\)

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

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

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

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

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

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

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

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

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

\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)

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)

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

Bài 7: Phân tích đa thức thành nhân tử

a) Ta có: \(a^2-b^2-2a+2b\)

\(=\left(a-b\right)\left(a+b\right)-2\left(a-b\right)\)

\(=\left(a-b\right)\left(a+b-2\right)\)

b) Ta có: \(3x-3y-5x\left(y-x\right)\)

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

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

c) Ta có: \(16-x^2+4xy-4y^2\)

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

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

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

d) Ta có: \(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

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

\(=\left(5-x-4y\right)\left(3x+2y+3\right)\)

e) Ta có: \(x^4+x^3+2x^2+x+1\)

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

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

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

f) Ta có: \(\left(x+3\right)^3+\left(x-3\right)^3\)

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

\(=2x\cdot\left[x^2+6x+9-\left(x^2-9\right)+x^2-6x+9\right]\)

\(=2x\cdot\left(2x^2+18-x^2+9\right)\)

\(=2x\cdot\left(x^2+27\right)\)

g) Ta có: \(9x^2-3xy+y-6x+1\)

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

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

\(=\left(3x-1\right)\left(3x-1-y\right)\)

h) Ta có: \(x^3-4x^2+12x-27\)

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

\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-x+9\right)\)

           Bài làm :

 \(\text{a)}9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)

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

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

\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)

 \(\text{b)}3x^4y^2+3x^3y^2+3xy^2+3y^2\)

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

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

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

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

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

 \(\text{c)}\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

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

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

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

\(d ) x^3+3x^2+3x+1-27z^3\)

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

a) 5xy ( x - y ) - 2x + 2y

= 5xy ( x - y ) - 2 ( x - y )

= ( x - y ) ( 5xy - 2 )

b) 6x-2y-x(y-3x)

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

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

c)  x² + 4x - xy-4y

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

( x + 4 ) ( x - y )

d) 3xy + 2z - 6y - xz 

= ( 3xy - 6y ) + ( 2z - xz )

= 3y ( x - 2 ) + z ( x - 2 )

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

a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)

b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)

c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)

d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)

11)

a,4-9x^2=0

(2-3x)(2+3x)=0

2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3

b,x^2 +x+1/4=0

(x+1/2)^2 =0

x+1/2=0

x=-1/2

c,2x(x-3)+(x-3)=0

(x-3)(2x+1)=0

x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2

d,3x(x-4)-x+4=0

3x(x-4)-(x-4)=0

(x-4)(3x-1)=0

x-4=0=>x=4 hoặc 3x-1=0=>x=1/3

e,x^3-1/9x=0

x(x^2-1/9)=0

x(x+1/3)(x-1/3)=0

x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3

f,(3x-y)^2-(x-y)^2 =0

(3x-y-x+y)(3x-y+x-y)=0

2x(4x-2y)=0

4x(2x-y)=0

x=0hoặc 2x-y=0=>x=y/2

thu thủy học lớp 9 chưa có long lone

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

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

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

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

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

d) y^2+y-x^2+x

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

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

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

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

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

f)9x^2-25y^2-6x+10y

= 

ko biết làm