ab(a-b)+bc(b-c)+ac(c-a) 
=ab(a-b)-bc[(a-b)+(c-a)] +ac(c-a) 
=ab(a-b) -bc(a-b) -bc(c-a) + ac(c-a) 
=b(a-c)(a-b) -c(a-c)(a-b) 
=(b-c)(a-c)(a-b) 

a² + 2ab + b² - ac - bc

= (a+b)² -c(a+b)

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

ab(a+b)+bc(b+c)+ca(c+a)+3abc

=(ab(a+b)+abc)+(bc(b+c)+abc)+(ca(c+a)+abc)

=ab(a+b+c)+bc(b+c+a)+ca(c+a+b)

=(a+b+c)(ab+bc+ca)

\(ab\left(x^2+y^2\right)-xy\left(a^2+b^2\right)\)

\(=abx^2+aby^2-a^2xy-b^2xy\)

\(=ax\left(bx-ay\right)+by\left(ay-bx\right)\)

\(=ax\left(bx-ay\right)-by\left(bx-ay\right)\)

\(\left(bx-ay\right)\left(ax-by\right)\)

hãy k nếu bạn thấy đây là câu tl đúng :)

Thanks 

\((4x-y)(a+b)(4x-y)(c-1)\)

\(=\left(4x-y\right)\left(4x-y\right)=\left(4x-y\right)^{1+1}=\left(4y-2\right)^2\)

\(=\left(a+b\right)\left(4x-y\right)^2\left(c-1\right)\)

(4x-y)(a+b)(4x-y)(c-1)

= ( 4x - y ) ( 4x - y ) = ( 4x - y ) ^{1 + 1} = ( 4y - 2 ) ²

= (a + b ) ( 4x - y )²  ( c - 1 )

Bài giải : 

(4x-y)(a+b)(4x-y)(c-1)

= ( 4x - y ) ( 4x - y ) = ( 4x - y ) ^{1 + 1} 

= ( 4y - 2 ) ²

= (a + b ) ( 4x - y )²  ( c - 1 )

\(=\left(3a-a^2\right)-\left(3b-ab\right)=\left(a-b\right)\left(3-a\right)\)

3a+3b-a²-ab

= 3a - a² + 3b - ab

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

= ( a+b)(3 - a )

3a + 3b - a² - ab

= 3a - a² + 3b - ab

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

= ( a + b )( 3 - a )