bạn giữ nguyên ab(a+b),mấy cái kia triển khai ra,xong rồi nhóm hạng tử thu gọn lại,dc nhân tử chung là (a+b),xong rồi đưa (a+b) lm nhân tử chung,bên trong thu gọn từ từ

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Còn lại bạn tự giải nha

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ai kết bạn mình cho

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

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

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

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

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

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

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

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

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

\(=\left(a^2b+ba^2+abc\right)+\left(b^2c+c^2b+abc\right)+\left(ac^2+ca^2\right)\)

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

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

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

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

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

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

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

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

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

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

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