\(ax^2-3axy+bx-3by\\ =x\left(ax+b\right)-3y\left(ax+b\right)\\ =\left(x-3y\right)\left(ax+b\right)\)

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

\(2ax^3+6ax^2+6ax+18a\\ =2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\)

\(10xy^2-5by^2+2ax-ab\\ =5y^2\left(2x-b\right)+a\left(2x-b\right)\\ =\left(5y^2+a\right)\left(2x-b\right)\)

\(ax-bx+cx-3a+3b-3c\\ =x\left(a-b+c\right)-3\left(a-b+c\right)\\ =\left(x-3\right)\left(a-b+c\right)\)

a) \(2a^2x-5by-5a^2y+2by\) 

 \(=3\left(\frac{2}{3}a^2x-\frac{5}{3}a^2y\right)-3by\) 

\(=3\left(\frac{2}{3}a^2x-\frac{5}{3}a^2y-by\right)\) 

Bài 2:

a) x(x - 3)- y(3 - x)

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

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

Thay x = \(\frac{1}{3}\); y = \(\frac{8}{3}\)vào (1)

Ta có: (\(\frac{1}{3}\)- 3)(\(\frac{1}{3}\)+ \(\frac{8}{3}\))

= \(\frac{-8}{3}\). 3

= -8

\(2ax-bx+3cx-2a+b-3c\\ =x\left(2a-b+3c\right)-\left(2a-b+3c\right)\\ =\left(x-1\right)\left(2a-b+3c\right)\)

\(ax-bx-2cx-2a+2b+4c\\ =x\left(a-b-2c\right)-2\left(a-b-2c\right)\\ =\left(x-2\right)\left(a-b-2c\right)\)

\(3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\)

\(ax^2-bx^2-2ax+2bx-3a+3b\\ =x^2\left(a-b\right)-2x\left(a-b\right)-3\left(a+b\right)\\ =\left(x^2-2x-3\right)\left(a+b\right)\\ =\left(x+1\right)\left(x-3\right)\left(a+b\right)\)

a, x(a - b) + (a - b)

= (x + 1)(a - b)

b, x(a + b) - a - b

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

= (x - 1)(a + b)

c, 10ax - 5ay - 2x + y

=  5a(2x - y) - (2x - y)

= (5a - 1)(2x - y)

d, 2a^2x - 5by - 5a^2y + 2bx

= 2x(a^2 + b) - 5y(b + a^2)

= (2a - 5y)(a^2 + b)

làm tiếp:

2ax² - bx² - 2ax +bx +4a-2b

= x²(2a-b) - x(2a-b) +2(2a-b)

=(2a-b)(x²-x+2)

1)\(x\left(a-b\right)+a-b=x\left(a-b\right)+\left(a-b\right)=\left(a-b\right)\left(a+x\right)\)

2)\(x\left(a+b\right)-a-b=x\left(a+b\right)-\left(a+b\right)=\left(a+b\right)\left(x-1\right)\)

3)\(10ax-5ay-2x+y=\left(10ax-5ay\right)-\left(2x-y\right)=5a\left(2x-y\right)-\left(2x-y\right)=\left(2x-y\right)\left(5a-1\right)\)

4)\(2a^2x-5by-5a^2y+2bx=\left(2a^2x-5a^2y\right)+\left(2bx-5by\right)=a^2\left(2x-5y\right)+b\left(2x-5y\right)=\left(2x-5y\right)\left(a^2+b\right)\)5)\(2ax^2-bx^2-2ax+bx+4a-2b=\left(2ax^2-bx^2\right)-\left(2ax-bx\right)+\left(4a-2b\right)=x^2\left(2a-b\right)-x\left(2a-b\right)+2\left(2a-b\right)\)\(=\left(2a-b\right)\left(x^2-x+2\right)\)

Bn viet bây à

Giúp mik đi

1, x(a-b)+a-b          2, x-y-a(x-y)            3, a(x+y)-x-y               4, x(a-b)-a+b         5, x²+xy-2x-2y            6, 10ax-5ay+2x-y

= x(a-b)+(a-b)           =(x-y)-a(x-y)             =a(x+y)-(x+y)             =x(a-b)-(a-b)         =(x²+xy)-(2x+2y)        =(10ax+2x)-(5ay+y)

=(a-b)(x+1)               =(x-y)(1-a)                =(x+y)(a-1)                 =(a-b)(x-1)            =x(x+y)-2(x+y)            =2x(5a+1)-y(5a+1)

                                                                                                                                  =(x+y)(x-2)                  =(5a+1)(2x-y)

7, 2a²x-5by-5a²y+2bx           8, 2ax²-bx²-2ax+bx+4a-2b             9, 2ax-bx+3cx-2a+b-3c                     10, ax-bx-2cx-2a+2b+4c

=(2a²x+2bx)-(5by+5a²y)           =(2ax²-bx²)-(2ax-bx)+(4a-2b)        =(2ax-2a)-(bx-b)+(3cx-3c)                  =(ax-2a)-(bx-2b)-(2cx-4c)

=2x(a²+b)-5y(b+a²)                   =x²(2a-b)-x(2a-b)+2(2a-b)             =2a(x-1)-b(x-1)+3c(x-1)                      =a(x-2)-b(x-2)-2c(x-2)

=(a²+b)(2x-5y)                           =(2a-b)(x²-x+2)                              =(x-1)(2a-b+3c)                                  =(x-2)(a-b-2c)

A = 4acx + 4bcx + 4ax + 4bx ( đã sửa '-' )

= 4x( ac + bc + a + b )

= 4x[ c( a + b ) + ( a + b ) ]

= 4x( a + b )( c + 1 )

B = ax - bx + cx - 3a + 3b - 3c

= x( a - b + c ) - 3( a - b + c )

= ( a - b + c )( x - 3 )

C = 2ax - bx + 3cx - 2a + b - 3c

= x( 2a - b + 3c ) - ( 2a - b + 3c )

= ( 2a - b + 3c )( x - 1 )

D = ax - bx - 2cx - 2a + 2b + 4c

= x( a - b - 2c ) - 2( a - b - 2c )

= ( a - b - 2c )( x - 2 )

E = 3ax² + 3bx² + ax + bx + 5a + 5b

= 3x²( a + b ) + x( a + b ) + 5( a + b )

= ( a + b )( 3x² + x + 5 )

F = ax² - bx² - 2ax + 2bx - 3a + 3b

= x²( a - b ) - 2x( a - b ) - 3( a - b )

= ( a - b )( x² - 2x - 3 )

= ( a - b )( x² + x - 3x - 3 )

= ( a - b )[ x( x + 1 ) - 3( x + 1 ) ]

= ( a - b )( x + 1 )( x - 3 )

A = 10ax - 5ay - 2x + y

= ( 10ax - 5ay ) - ( 2x - y )

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

= ( 2x - y )( 5a - 1 )

B = 2x² - 6xy + 5x - 15y 

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

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

C = ax² - 3axy + bx - 3by

= ( ax² + bx ) - ( 3axy + 3by )

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

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

D = 2ax³ + 6ax² + 6ax + 18a

= 2ax²( x + 3 ) + 6a( x + 3 )

= ( x + 3 )( 2ax² + 6a )

= ( x + 3 )2a( x² + 3 )

E = 5x²y + 5xy² + a²x + a²y ( đã sửa 1 dấu '-' )

= 5xy( x + y ) + a²( x + y )

= ( x + y )( 5xy + a² )

F = 10xy² - 5by² + 2a²x - aby ( xem lại đề chứ không phân tích được :)) )

a) Ta có: \(4\left(2-x\right)^2+xy-2y\)

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

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

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

b) Ta có: \(3a^2x-3a^2y+abx-aby\)

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

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

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

c) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)

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

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

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

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

d) Ta có: \(2ax^3+6ax^2+6ax+18a\)

\(=2ax^2\left(x+3\right)+6a\left(x+3\right)\)

\(=\left(x+3\right)\left(2ax^3+6a\right)\)

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

e) Ta có: \(x^2y-xy^2-3x+3y\)

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

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