If the real numbers a and B satisfy that the root a plus 1 plus absolute value b minus 1 equals 0, then the 2010 power of a plus the 2011 power of B is equal to

If the real numbers a and B satisfy that the root a plus 1 plus absolute value b minus 1 equals 0, then the 2010 power of a plus the 2011 power of B is equal to

According to the question, a = - 1, B = 1
So the 2010 power of a plus the 2011 power of B = 1 + 1 = 2

How much is the root difference of the number five plus six in 2006?

Root number [+ 6] = (5) root sign (6)
=(6-5) ^ (2006) * (root 5 + root 6)
=Root 5 + root 6

The power of 2007 enclosed by root 2 minus root 3 × root 2 minus root 3 equals to the power of 2008?

Root two minus root three enclosed 4015 power

Calculate: (2-radix-3) to the power of 2006 times the power of (2 + root 3) to the power of 2007 - (- root 2) to the power of 0=

The original formula = (2 - √ 3) ^ 2006 * (2 + √ 3) ^ 2007 - (- √ 2) ^ 0
=[(2-√3)(2+√3)]^2006*(2+√3)-1
=1^2006*(2+√3)-1
=1+√3

(root 3-root 2) 2006 power X (root 3 + root 2) 2007 power =?

(root 3-root 2) 2006 power X (root 3 + root 2) 2007 power
=[(√ 3 - √ 2) (√ 3 + √ 2)] 2006 Power (√ 3 + √ 2)
=(3-2) the power of 2006 (√ 3 + √ 2)
=√3+√2

Calculate the zero power of the minus one-half of the absolute value minus the root sign four plus (π - 4)

.
|-1/2|-√4+(π-4)^0
=1/2-2+1
=-1/2

If ABC is a rational number and satisfies the equation a + B radical 5 + C radical 7 = 2-radical 5 + 3 root sign 7, try to calculate the value of (A-C) 2012 power + B's 2013 power

A + B radical 5 + C radical 7 = 2-radical 5 + 3 radical 7
(A-2) + (b root 5 + root 5) + (C root 7-3 root 7) = 0
Because ABC is a rational number
So a = 2, B = - 1, C = 3
（a-c）^2012+b^2013=0

It is known that the three side lengths a, B and C of △ ABC satisfy a−1+|b−1|+(c− 2) 2 = 0, then △ ABC must be______ Triangle

∵ △ the three side lengths a, B and C of ABC are satisfied
a−1+|b−1|+(c−
2)2＝0，
∴a-1=0，b-1=0，c-
2=0，
∴a=1，b=1，c=
2．
∵a2+b2=c2，
The △ ABC must be an isosceles right triangle

Let A.B.C be the three sides of the triangle ABC, and satisfy the relation a to the power of 2 + the power of B + the power of C + 10 = 2A + 4B + 2 times the root sign 5 times C, and judge the shape of the triangle

∵a²+b²+c²+10=2a+4b+2√5c ∴(a-1²)+(b-2)²+(c-√5)²=0
If a = 1, B = 2, C = √ 5, then a? + B? = 1 + 4 = 5 = C? ﹤ triangle ABC is a right triangle with angle c as its right angle

(2 times root 2-3) 2011 power × (2 times root 2 + 3) 2012 power, - 3 and + 3 are not in root 2!

The original formula = [(2 roots 2-3) * (2 roots 2 + 3)] ^ 2011 * (2 roots 2 + 3)
=(- 1) ^ 2011 * (2 roots, 2 + 3)
=-2-3