The three sides of ABC are a, b, and c, which are simplified to a b c b b a c a b c

Solution: Because the sum of any two sides of the triangle is greater than the third side, the difference between any two sides is less than the third side, so a 0 , b 0 , c 0

a-b-c 0 b-a-c 0 a+b+c 0So, the original formula =-(a-b-c)-(b-a-c)-(a+b+c)-a+b+c-a+c-a-b-c

a-b+c Description: The key to removing the absolute value is to see whether the value of the formula in the absolute value is positive or negative. It is positive and gets itself after removing the absolute value sign, and if it is negative, it is the opposite of the original formula after removing the absolute value sign.

a+b-c|-|b-a-c|-|a-b-c|a+b-c-<-b-a-c)-a+b-c+b-a-c+a-b-c

a+b-3c

PS: Sum of any 2 sides of a triangle 3rd side.

Any difference between the two sides of the 3rd side.

Because the three sides of ABC are A, B, and C.

So a+b is greater than c, b-a is less than c, and a-b is less than c

So |a+b-c|-|b-a-c|-|a-b-c|=a+b-c-（a+c-b)-(b+c-a)=a+b-3c

The original formula a+b-c-(a+c-b)-(b+c-a)=a+b-3c because in a triangle, there is a b

Greater than c, a c greater than b, b c greater than a

It is then calculated according to the meaning of the absolute value

The three-sided long collision calendar of ABC is a, b, and c, and the sum of the two sides must be greater than the third side, then a-b+c 0, and the line is large a-b-c 0, |a-b+c|-|a-b-c|=a-b+c+a-b-c=2a-2b．

Therefore, the answer file is vertically arguing: 2a-2b

So the two sides and the third side are larger.

So |a-b-c|+|b-c-a|

b+c-a+a+c-b

The sum of the two sides of the triangle is greater than the third side.

The banquet collapsed on the mountain with a+b-c>0

b-a-c<0

So the original formula = a+b-c+b-a-c=2b-2c

Because the sum of the two sides is greater than the third side, a+b c, b-a-c 0

Original formula = a+b-c-[-b-a-c)].

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

a+b-c+b-a-c

The sum of the two sides of the triangle is greater than the third side.

So a+b-c>0

b-a-c<0

So the original formula = a+b-c+b-a-c=2b-2c

For a triangle, the sum of any two sides is greater than the third side.

Thus a+b>c a+c>b

a+b-c|-|b-a-c|=|a+b-c|-|b-（a+c）|=a+b-c-a-c+b=2（b-c)

Ask questions in a timely manner.

The sum of the two sides of the triangle must be greater than the third side.

a+b-c|-|b-a-c|=a+b-c-(a+c-b)=a+b-c-a-c+b=2b-2c

The answer should be a+b+c

a-b-c|+ b-c-a|+ c-a-b|=|a-(b+c)|+b-(c+a)|+c-(a+b)|for the sum of the two sides is greater than the third side, and if a does not know whether it is right or not.

Because it's a triangle, the sum of the two sides is greater than the third side.

a+b>c ,a+c>b ,b+c>a

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

a+b+c

This problem is mainly simplified by using the sum of the two sides of the triangle to be greater than the third side.

a-b-c=a-(b+c)<0 , so |a-b-c|=b+c-a;

In the same way: |b-c-a|=c+a-b |c-a-b|=a+b-cSo:|a-b-c|+ b-c-a|+ c-a-b|=a+b+c

The sum of the two sides of the triangle is greater than the third side.

So a+b-c>0

b-a-c<0

c-a-b<0

Original formula =-a+b+c+b-a-c+c-a-b=-3a+b+c

The sum of the two sides of the triangle is greater than the third side.

So A + B > C, A + C > B, B + C > A

So a - b - c < 0 , b - c - a < 0 , c -a - b < 0

So |a-b-c| +b-c-a| +c-a-b|= b + c - a + a + c - b + a + b - c

a + b + c

Solution: a, b, c are the three sides of a triangle.

So b+c>a

a-b-c<0

a+c>b

b-a-c<0

c+a>b

c+a-b>0

then |a-b-c|+|b-c-a|+|c+a-b|=-a+b+c-b+c+a+a+c-b

3c+a-b

The three sides of ABC are A, B, and C

then there is a=b+c-a+c+a-b+a+b-c=a+b+c