Given that the absolute values of A-B + 3 and a + B-5 are opposite numbers, find the value of a + B?

Given that the absolute values of A-B + 3 and a + B-5 are opposite numbers, find the value of a + B?

In fact, it is not difficult to say that the number under the root sign is ≥ 0, and the absolute value is also ≥ 0. It is said that they are opposite numbers to each other, so the addition of two non negative numbers is zero, so they are respectively zero, so A-B + 3 = 0, a + B-5 = 0, a = 1, B = 4, so a + B = 5

If the absolute value of a number is root 3, then the number is

Root sign 3

|-The 2011 power of 3 | + (- 1) × (π - 3) to the power of 0-3 and 27 + (1 / 2) to the power of - 2

|-The 2011 power of 3 | + (- 1) × (π - 3) to the power of 0-3 and 27 + (1 / 2) to the power of - 2
=3+（-1）×1-3+2²
=3-1-3+4
=3

Calculation: - the root of the third power - 1 / 27 The root sign of the third power is 8 × the root sign of the third power - 1 / 64; find the value of X in the following formulas: 8x 2 + 1 = 0

- third power root sign - 1 / 27
=-Third power root (- 1 / 3) 3
=-(- 1 / 3)
=1 / 3
The root of the third power is 8 × the root of the third power - 1 / 64
=Root sign of the third power [8 × (- 1 / 64)]
=Third power root (- 1 / 8)
=-1 / 3
8x³+1=0
8x³=-1
X 3 = - 1 / 8
X = - 1 / 2

Calculation: cubic root number 27 - root number 144 third power root number 216 * cubic root number 512 -Is a minus sign, * is a multiplier sign

Original formula = 3-6 / 12 × 8 = - 1

Calculation: the third power of root 3x is 27

√（27 / 3 x ³）
= √（27 · 3 x / 3 ² x ⁴）
= √（9 ² x ）/ 3 x ²
= 9√x / 3 x ²
= 3√x / x ²
Premise: x > 0

The square of - 3 + (- 1) multiplied by the 0 th power of (π - 3) - the third power of root 27 + (1 / 2) the negative second power

The square of - 3 + (- 1) multiplied by the 0 th power of (π - 3) - the third power of root 27 + (1 / 2) the negative second power
=3-1-3+4
=3

The third is the third of the fifth root 3 times root number 12-4 times root number 1-2 times root number 27

The second power of (Radix 5 - (2 / 5 of radix))
=5-2 root sign 2 + 2 / 5
=5.4-2 root sign 2
3 times root number 12-4 times root number 1-2 times root number 27
=6 roots 3-4 / 3 roots 3-6 roots 3
=-4 / 3 root sign 3

Calculate the absolute value of (π - 2009) to the power 0 + root 12 + root 3-2

(π-2009)^0=1
√12=2√3
Then the original formula = 1 + 2 √ 3 + √ 3-2 = 3 √ 3-1
Because of 3 √ 3 ≠ 1
The absolute value is 3 √ 3-1

A + 2 + | B-1 | = 0, then the value of (a + b) 2009 is___ .

According to the meaning of the title:
a+2=0
b-1=0 ，
The solution is as follows:
a=-2
b=1 ，
Then the original formula = (- 2 + 1) 2009 = - 1
So the answer is: - 1