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

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

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

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

A = 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 )

ax²-5x²-ax+5x+a-5

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

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

cậu ghi sai đề rồi phải là

3ax²+3bx²+ax+bx+5a+5b

=3x^2(a+b)+x(a+b)+5(a+b)

=(a+b)(3x^2+x+5)

\(1,3ax^2+3bx^2+ax+bx+5a+5b\)

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

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

(b-a-5)*(b-a)

a²+b²-5b-2ab+5a
= (b²-2ab+a²)-5(b-a)
= (b-a)²-5(b-a)
= (b-a)(b-a-5)

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: \(x^2y-xy^2-3x+3y\)

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

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

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

\(5a^2-5b^2-20a+20b\)

\(=5\left(a^2-b^2\right)-20\left(a-b\right)\)

\(=5\left(a-b\right)\left(a+b\right)-20\left(a-b\right)\)

\(=\left[5\left(a+b\right)-20\right]\left(a-b\right)\)

\(=\left(5a+5b-20\right)\left(a-b\right)\)

\(5a^2-5b^2-20a+20=-5.\left(b-a+2\right).\left(b+a-2\right)\)