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25 tháng 7 2016

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

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

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

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

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

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

\(=\left(a+b\right)\text{[}a\left(b+c\right)-c\left(b+c\right)\)

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

7 tháng 10 2018

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

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

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

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

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

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

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

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

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

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

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

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

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

Tham khảo nhé~

8 tháng 10 2018

thank you

14 tháng 3 2019

Ta có ab(a – b) + bc(b – c) + ca(c – a)

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

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

= (a – b)(ab – bc) + (a – c)(bc – ac)

= b(a – b)(a – c) – c(a – c)(a – b)

= (a – b)(a – c)(b – c)

Đáp án cần chọn là: A

17 tháng 9 2019

Thêm bớt abc vào M ta có

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

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

=(a + b + c)(ab – bc + ac)

Đáp án cần chọn là: D

14 tháng 10 2018

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)

14 tháng 10 2018

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

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

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

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

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

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

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

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

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

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

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

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

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

c: Ta có: \(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)

\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)

\(=a^4-2a^3b+2ab^3-b^4\)

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

\(=\left(a-b\right)^3\cdot\left(a+b\right)\)

31 tháng 8 2017

ai biết trả lời nhanh hộ mình nha! Mình k đúng cho!

31 tháng 8 2017

Co P=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) 

2 tháng 9 2017

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

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

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

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

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

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

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

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