ta có: ab(a + b) + bc(b + c) + ac(a + c) + 3abc 

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

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

= (a + b + c)(ab + bc + ca) 

\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)

\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)\)

\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a\right)\)

\(=\left(a+b+c\right)\left(ab+bc\right)+ca\left(c+a\right)\)

\(=b.\left(a+b+c\right)\left(a+c\right)+ca\left(c+a\right)\)

\(=\left(a+c\right)\left[b.\left(a+b+c\right)+ca\right]\)

\(=\left(a+c\right)\left(ab+b^2+bc+ca\right)\)

\(=\left(a+c\right)\left[a\left(b+c\right)+b\left(b+c\right)\right]\)

\(=\left(a+c\right)\left(b+c\right)\left(a+b\right)\)

\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc\)

\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)+abc\)

\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a+b\right)\)

\(=\left(a+b+c\right)\left(ab+bc+ac\right)\)

Tham khảo nhé~

thank you

Ta có: \(D=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc\)

\(=a^2b+ab^2+b^2c+bc^2+ac^2+a^2c+3abc\)

\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

\(=a^2b+ab^2-b^2c-bc^2-ac^2+a^2c\)

\(=a^2\left(b+c\right)+a\left(b-c\right)\left(b+c\right)-bc\left(b+c\right)\)

\(=\left(b+c\right)\left(a^2+ab-ac-bc\right)\)

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

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

Ta có b + c = (a + b) + (c – a) nên

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 + b)(a – c) – c(c – a)(b + a)

= (a + b)(a – c)(b + c)

Đáp án cần chọn là: B

2: =abc-bc-ab-ac+a+b+c-1

=bc(a-1)-ab+b-ac+c+a-1

=bc(a-1)-b(a-1)-c(a-1)+(a-1)

=(a-1)(bc-b-c+1)

=(a-1)(b-1)(c-1)

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