Question

phân tích thành nhân tử

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

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

 

Answer 1

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)

Answer 2

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

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

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

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

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

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

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

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

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

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