If a, b, and c are all integers, and a b 2001 c a 2000 1, find the value of a c c b b a

2001 and 2000 should be indices.

1) This problem can be discussed in different situations: for example, A>B>C, A>B>C can be solved.

2) Usually tell you that it is an integer, that is, prompt us to say that their difference and sum, product and so on are integers.

That is, a-b, c-a, and c-b are also integers.

And the absolute value cannot be negative.

That is, a-b = 0, c-a = 1 or -1, that is, a = b, c-b is c-a, the absolute value is the same, all are 1

a-b=1 or -1 c-a=0.

The answer is 2 Some symbols won't hit, so that's the basics.

Solution: a, b, and c are all integers, and |a-b|2001+|c-a|2000=1

a-b|=1，|c-a|=0 or |a-b|=0，|c-a|=1 when|a-b|=1，|c-a|=0, c=a, a=b 1, so |a-c|+|c-b|+|b-a|=|a-c|+|a-b|+|b-a|=0+1+1=2；

When|a-b|=0，|c-a|=1

a=b, so |a-c|+|c-b|+|b-a|=|a-c|+|c-a|+|b-a|=1+1+0=2；

Comprehensively, it can be seen that: |a-c|+|c-b|+|b-a|The value of 2 is b

a-b|,|c-a|The possible values are 0, 1, and 2,..

So:|a-b|2001+|c-a|The possible values of 2000 are: 0, 2000, 2001, 4001, and cannot be equal to 1

a-b|2001+|c-a|2000=1, this equation does not hold.

The reason is that a, b and c are all integers.

a, b, and c are all integers, and |a-b|+|a-c|=1|a-b|=1 or |a-c|=1

When|a-b|=1, |a-c|=0,a=c,|c-b|=|a-b|=1

c﹣a| +c-b|+|b-a|=0+1+1=2 dang|a-c|=1, |a-b|=0,a=b,|c-b|=|c-a|=1

c﹣a| +c-b|+|b-a|=1+1+0=2 Therefore, if a, b, and c are all integers of the dry cave of the feast, and |a-b|+|a-c|=1, then |c﹣a| +c-b|+|b-a|The constant is equal to 2

It is known that a, b, and c are all integers, so defeating a-b and c-a are also integers; And there is a rolling limb: a-b + c-a =1 , you get:

a-b =0, c-a =1 or a-b =1, c-a =0When a-b = 0 and c-a = 1, then c-a + a-b + b-c = c-a

If abc is an integer with an implicit |a-b|+|c-a|=0a-b=0;

c-a=0;

stupid or a=b=c;

Ask for ita-c|+|c-b|+b-a|value.

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A, b, and c are all integers, and |a-b| 3 +|c-a|2 = 1, cover filial piety.

a, b, and c have two equal numbers, and if they are set to a=b, then |c-a|=1, c=a+1 or c=a-1, |a-c|=|a-a-1|=1 or |a-c|=|a-a+1|=1，|a-c|+|c-b|+|b-a|=1+1=2．

a, mu b, land c are all integers, and |a-b| 2012 +|c-a|2012 =1, out a, b, c two numbers are equal, you may wish to set a=b, then |c-a|=1, c=a+1 or c=a-1, |b-c|=|a-a-1|=1 or |b-c|=|a-a+1|=1, Chang Xingsen.

c-a|+|a-b|+|b-c|=1+1=2．

Because a, b, and c are all integers, and |a-b|>0，|c-a|>0, and |a-b|+|c-a|=1

Therefore, there must be a result of 0 and a demerit of 1

Assumptions|a-b|=0

c-a|1, then a=b, c=a+1

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

a-a-1|+|a+1-a|+|a-a|=2 Hypothesis|a-b|=1

c-a|0, then a=b+1, c=a

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

a-a|+|b-b-1|+|b-b-1|=2