解:∵ a,b,c是三角形的三边长,而三角形和两边之和大于第三边,两边之差小于第三边
即:\(\displaystyle \begin{array}{l}a+b>c\\a-b<c\end{array}\)
∴ 原式\(\displaystyle =\left| {b+a-c} \right|+\left| {a-b-c} \right|-\left| {b-a-c} \right|\)
\(\displaystyle =b+a-c-a+b+c+b-a-c\)
\(\displaystyle =3b-a-c\)
