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)

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

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

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

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

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

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

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

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

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

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

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

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) 

bạn viết đề sai kìa bạn

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Tham khảo nhé~

thank you

X=ab(a+b)+bc(b+c)+ca(c+a)+2abc

=a²b+ab²+b²c+bc²+c²a+a²c+2abc

=b(a+c)²+b²(a+c)+ac(a+c)+2abc

=[b(a+c)+b²+ac](a+c)+2abc

=(ab+bc+b²+ac)(a+c)+2abc

=[b(c+b)+a(b+c)](a+c)+2abc

=[(b+a)(b+c)](a+c)+2abc

rồi j nữa, tịt rồi

giúp đi

X=ab(a+b)+bc(b+c)+ca(c+a)+2abc

=a²b+ab²+b²c+bc²+ac²+a²c+2abc

=a(ab+b²+c²+ac+2bc)+bc(b+c)

=a[(b+c)²+a(b+c)]+bc(b+c)

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

=a(b+c)(a+b+c+bc)
Xin hết

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

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

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

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

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

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

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

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

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

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

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

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

       Đến đây thì mk chịu