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

= a²b + abc +a²c + ab² + b²c + abc + abc + bc² + ac²

= (a²b + 2abc + bc²) + (ac² + a²c) + (ab² + b²c)

= b(a² + 2ac + c²) + ac(c + a) + b²(a + c)

= b(a + c) + ac(a + c) + b²(a + c)

= (a + c)[b(a + c) + ac + b²]

= (a + c)(ab + bc + ac + b²)

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

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

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

\(=a^2b+abc+a^2c+b^2a+b^2c+abc+abc+c^2b+c^2a-abc\)

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

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

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

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

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

= abc - bc - 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)

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

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

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

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

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

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

\(A=\left(a+b+c\right).\left(bc+ca+ab\right)-abc\\ =abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc-abc\\ =\left(b^2c+bc^2\right)+\left(a^2b+a^2c\right)+\left(ac^2+abc\right)+\left(ab^2+abc\right)\\ =bc\left(b+c\right)+a^2\left(b+c\right)+ac\left(b+c\right)+ab\left(b+c\right)\\ =\left(b+c\right)\left(bc+a^2+ac+ab\right)\\ =\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]=\left(b+c\right)\left(a+c\right)\left(a+b\right)\)

(a + b + c)(bc + ca + ab) − abc

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

=(a + b)(bc + ca + ab)+ abc + c²(a + b) − abc

=(a + b)(bc + ca + ab + c²)

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

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

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

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

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

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

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

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

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

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)