Known as | a || a+|b| b+|c| C = - 1, try to find ab |ab|+bc |bc|+ca |ca|+abc |The value of ABC |

Known as | a || a+|b| b+|c| C = - 1, try to find ab |ab|+bc |bc|+ca |ca|+abc |The value of ABC |

It can be concluded that there are two negative numbers and one positive number in a, B and C,
① If a < 0, B < 0, C > 0, then AB > 0, BC < 0, CA < 0, ABC > 0,
The original formula = 1-1-1 + 1 = 0;
② If a < 0, b > 0, C < 0, then AB < 0, BC < 0, CA > 0, ABC > 0,
The original formula = - 1-1 + 1 + 1 = 0;
In other cases, it is deduced that there are two positive numbers and two negative numbers in AB, BC, AC and ABC,
So: ab
|ab|+bc
|bc|+ca
|ca|+abc
|abc|=0．

Given that the absolute value of a + 1 + the 2nd power of (b-2) = 0, calculate the value of (a + b) to the power of 2010 + the power of 2011 of A

If there is an absolute sum of 0, then it is not equal to 0
So both are equal to zero
So a + 1 = 0, B-2 = 0
a=-1,b=2
a+b=1
So the original formula = 1 to the power of 2010 + (- 1) to the power of 2011
=1-1
=0

If a, B, C are integers and the 2011 power of the absolute value of A-B + the 2013 power of the absolute value of C-A = 1 Find the absolute value of A-B + the absolute value of b-c

The 2011 power of absolute value of A-B + 2013 power of absolute value of C-A = 1
therefore
1.a-b=0,c-a=1
c-b=1
Absolute value of A-B + absolute value of B-C
=0+1=1
Two
a-b=0,c-a=-1
c-b=-1
Absolute value of A-B + absolute value of B-C
=0+1=1
Three
a-b=1,c-a=0
c-b=1
Absolute value of A-B + absolute value of B-C
=1+0=1
Four
a-b=-1,c-a=0
c-b=-1
Absolute value of A-B + absolute value of B-C
=1+0=1
therefore
Original formula = 1

Given that the absolute value of part a of a + the absolute value of B of Part B + the absolute value of C of Part C = 1, calculate the 1993 power of the absolute value of ABC of fraction a and the absolute value of ab BC times the absolute value of BC, AC times the absolute value of Ca, and the value of AB)

When a number is greater than or equal to zero, its absolute value is equal to itself; when a number is less than zero, its absolute value is equal to its opposite number. Since the absolute value of a part of a + the absolute value of B part + the absolute value of C part of C = 1, then there are two cases: 1. - 1 + 1 + 1 = 1. In this case, the absolute value of ABC fraction of ABC = 1ab

If ABC is an integer, and | A-B | 2013 power + | C-A | 2012 power = 1, what is | a-c | + | C-B | + | B-A |

a. B and C are integers
|The 2013 power of A-B | and the 2012 power of | C-A | is equal to 1
|A-B | is an integer to the power of 2013
|The power of C-A | is an integer and is greater than or equal to 0
So there are only two possibilities
1,a-b=0,|c-a|=1
|a-b|+|b-c|+|c-a|=2
2,c-a=0,|a-b|=1
|a-b|+|b-c|+|c-a|=2

A / | a | + | B | / B + C / | C | = 1, find the value of (| ABC |) / ABC ﹤ (BC / | ab | × AC | × AB / | AC |

The key to this topic is to consider how to remove the sign of absolute value. From the known a /

Given a / | a | + | B | / B + | C | / C = 1, find the value of (| ABC | / ABC) to the power of 2003 divided by the value of (BC / | AC | AC / | BC | * AB / | AC |

From the first formula, a / | a | a |, | B | / B, | C | / C when a, B, C is greater than 0, the value is 1. If the result is 1, only two of a, B and C are positive and one is negative. (a, B, C can not take 0, which is meaningless)
So ABC

Known: ab (a + b) of the negative power = 1, BC (B + C) of the negative first power = 2, AC (a + C) of the negative first power = 3, try to find the value of ABC (AB + Ba + AC)

ab(a+b)^(-1)=1
bc(b+c)^(-1)=2
ac(a+c)^(-1)=3
therefore
a+b/ab=1 b+c/bc=1/2 a+c/ac=1/3
therefore
1/a+1/b=1
1/b+1/c=1/2
1/a+1/c=1/3
So it's solved
a=12/5
b=12/7
c=-12
Namely
abc/a+b+c=144/23

In △ ABC, BC = a, AC = B, ab = C, and satisfy A4 + B4 + 1 2C4 = a2c2 + b2c2. Try to determine the shape of △ ABC

a4+b4+1
The deformation of 2C4 = a2c2 + b2c2 is as follows:
a4+b4+1
2c4-a2c2-b2c2=0，
∴（a4-a2c2+1
4c4）+（b4-b2c2+1
4c2）=0，
∴(a2−1
2c2) 2+(b2−1
2c2)2=0，
∴a=b，
a2+b2=c2，
So △ ABC is an isosceles right triangle

If the three sides a, B, C of △ ABC satisfy the condition that the second power of a - the second power of C = AB BC, then △ ABC is

a^2-c^2=ab-bc
(a+c)(a-c)=b(a-c)
(a+c)(a-c)-b(a-c)=0
(a+b-c)(a-c)=0
A=c
So it's an isosceles triangle