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

\(N=x^2-3x-2xy+y^2+3y=\left(x-y\right)\left(x-y-3\right)\)\(K=2xy+3z+6y+xz=\left(x+3\right)\left(2y+z\right)\)

M= x²-5x+xy-5y= x(x-5)+y(x-5)=(x-5)(x+y)

N= x²-3x-2xy+y²+3y=(x-y)²-3(x-y)=(x-y)(x-y-3)

K= 2xy+3z+6y+xz=2y(x+3)+z(x+3)=(x+3)(2y+z)

a) \(6x-6y=6\left(x-y\right)\)

b)\(2xy+3x+6y+xz\)

\(=\left(2xy+xz\right)+\left(6y+3z\right)\)

\(=x\left(2y+z\right)+3\left(2y+z\right)\)

\(=\left(2y+z\right)\left(x+3\right)\)

c)\(x^2+6x+9-y^2\)

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

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

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

d) \(9x-x^3\)

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

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

e)\(x^2-xy+x-y\)

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

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

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

a, 6x - 6y = 6( x-y )

b, 2xy + 3z + 6y + xz

  = ( 2xy + 6y ) + ( 3z + xz )

  = 2y( x + 3 ) + z ( 3 + x )

  = 2y( 3 + x ) + z ( 3 + x )

  = ( 3 + x ) ( 2y + z )

c, x² + 6x + 9 - y² = ( x² + 6x + 9 ) - y²

                             = ( x + 3 )² - y²

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

d , 9x - x³ = x ( 9 - x² )

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

e, x² - xy + x - y =( x ² - xy ) + ( x - y )

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

                          = ( x - y ) ( x + 1 )

\(a,6x-6y=6\left(x-y\right)\)

\(b.2xy+3z+6y+xz\)

\(2y\left(x+3\right)+z\left(x+3\right)\)

\(\left(2y+z\right)\left(x+3\right)\)

\(c,x^2+6x+9-y^2\)

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

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

\(d,9x-x^3\)

\(x\left(9-x^2\right)\)

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

\(e,x^2-xy+x-y\)

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

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

a) Cách 1.

Ta có 2xy + 3z + 6y + xz = (2xy + xz) + (3z + 6y)

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

Cách 2.

Ta có 2xy + 3z + 6y + xz = (2x1/ + 6y) + (3z + xz)

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

b) Biến đổi được a 4   -   9 rt 3   +   a 2 -9a = (a- 9)a( a 2  +1).

c) Biến đổi được 3 x 2  + 5y - 3xy + (-5x) = (x - y)(3x - 5).

d) Biến đổi được  x 2  - (a + b)x + ab = (x- a)(x - b).

e) Ta có 4 x 2 - 4xy + y 2   –   9 t 2 =  ( 2 x   -   y ) 2   -   ( 3 t ) 2

= (2x - y - 3t )(2x - y + 31).

g) Ta có  x 3   -   3 x 2 y   +   3 xy 2   -   y 3   -   z 3

= ( x   -   y ) 3   -   z 3 = (x - y - z)( x 2   +   y 2   +   z 2  - 2xy + xz - yz).

h) Ta có x 2   -   y 2 + 8x + 6y+ 7 = ( x 2  +8x + 16) - ( y 2  - 6y+ 9)

= ( x   +   4 ) 2   - ( y - 3 ) 2  =(x-y + 7)(x + y + l).

a) x² - 6x +9 = (x-3)²  

b) x² - 64 = (x-8)(x+8)

c) 2xy+3z+6y+xz = (2xy+xz)=(3z+6y)= x(2y+z) + 3(2y+z)=(2y+z)(x+3)

d) 5x²+5xy-x-y = 5x(x+y)-(x+y) = (x+y)(5x-1)

e) x² - xy+ x-y = x(x-y)+(x-y) = (x-y)(x+1)

CHÚC BN HỌC TỐT

ai jup mjk s

bn xem lại đề xem hình như 3zy phải là 3z chứ 

3zy bạn ơi

Có 2xy + 3zy +6y + xz = 2xy + xz + 3zy + 6y = x(2y+z) + 3y(z+2y) =(2y+z)(x+3y)

bạn ơi chổ\(3xy+6y\)

nếu rut ra sẽ là \(3y\left(x+1\right)\)

xin lỗi là 3y(x+2)

Ý a có rì đó sai sai nha bn 

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

a/ 5x²-x+5xy-y

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

= (5x-1) (x+y)

b) 2xz + 3z + 6y + xz 

= ko bt làm

a; x(5x-1)+y(5x-1)

=>( x+y).(5x-1)

b; có ghi sai đề ko vậy ; toàn x vs z thế