Ta có:
\(S=-\left(a-b-c\right)+\left(-c+b+a\right)-\left(a+b\right)=-a+b+c-c+b+a-a-b=b-a\)
\(\text{Suy ra:}\) \(\left|S\right|=a-b\) (Vì a > b)
Vậy: |S| = a - b
S=-(a-b-c)+(-c+b+a)-(a+b)
S=-a+b+c+(-c)+b+a-a-b
S=b -a
TH1:
\(\left|S\right|\)=b-a
TH 2:
\(\left|S\right|\)=-(a-b)=-b+a=a-b