\(a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)\)
\(=ab-ac-ba-bc+ca-cb=-2bc\)
a(b-c)-b(a+c)+c(a-b)=ab-ab-bc-ac+ac-bc=-2bc
\(a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)\)
\(=ab-ac-ba-bc+ca-cb\)
\(=\left(ab-ba\right)+\left(-ac+ca\right)+\left(-bc-cb\right)\)
\(=0+0-2bc\)
\(=-2bc\)
Vậy \(a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)=-2bc\).
Học tốt
a( b - c ) - b( a + c ) + c( a - b )
= ab - ac - ab - bc + ac - bc
= ( ab - ab ) + ( ac - ac ) + ( -bc - bc )
= -2bc ( đpcm )
Bài làm :
Ta có :
\(a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)\)
\(=ab-ac-ba-bc+ca-cb\)
\(=\left(ab-ba\right)+\left(-ac+ca\right)+\left(-bc-cb\right)\)
\(=0+0-2bc\)
\(=-2bc\)
=> Điều phải chứng minh