If M = 2011 / (Radix 2011-1), then the quintic power of M - the fourth power of 2m - the cubic power of 2011m =?

If M = 2011 / (Radix 2011-1), then the quintic power of M - the fourth power of 2m - the cubic power of 2011m =?

The answer is the fourth power of 2011 / the fifth power of (root number 2011-1)

The 2010 power of (4 + Radix 15) is the 2011 power of (4-radix-15)

=(4 + √ 15) to the power of 2010 × (4 - √ 15) to the power of 2010 × (4 - √ 15)
=The 2010 power of [(4 + √ 15) × (4 - √ 15)] is × (4 - √ 15)
=(16-15) to the power of 2010 × (4 - √ 15)
=The 2010 power of 1 × (4 - √ 15)
=4-√15

The 2007 power of (2 + Radix 5) is multiplied by the 2008 power of (2-radix 5)

The 2007 power of (2 + Radix 5) is multiplied by the 2008 power of (2-radix 5)
= the 2007 power of (2 + Radix 5) times the 2008 power of (Radix 5-2)
= ((Radix 5-2) (2 + Radix 5)) multiplied by (Radix 5-2)
= 1 to the power of 2007 (Radix 5-2)
= radical 5-2

The 2007 power of the difference of 3 minus 2 under the root sign multiplied by the 2008 power of the difference of 3 minus 2 under the root sign is equal to? It's not that the formula can't be listed, but that I can't type the root, the power of 2007 and the power of 2008

To the power of 2007 times the power of 2008
=1*1
=1

The 2008 power of (root 3 + 1) - 2 (root 3 + 1) is 2007 power - 2 (root 3 + 1) is 2006 power + 2008

The power of 2008 of (root 3 + 1) - 2 of 2 (root 3 + 1) and the power of 2006 of - 2 (root 3 + 1) + 2008 = (root 3 + 1) of 2006 * [(root 3 + 1) - 2] + 2008 = (root 3 + 1) of 2006 * [4 + 2 root 3-2 root 3-2] + 2008 = (root 3 + 1) of 2006 Power * 0 + 2008 = 20

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

Separate and change positions
We get the 2006 power of (root 2 - root 3) * (root 2 + root 3) plus (root 2 + root 3)
The result is
Radical 2 + radical 3 + 1

Calculate: (1 * 3-1 / 5) * (1 / 5) to the power of - 2 × | - 1 / 3 | + (1-radical 3) to the power of 0 + (- 0.25) to the power of 2007 * 4 to the power of 2008

(1*3-1/5)*(1/5)^(-2)÷|-1/3|+(1-√3)^0+(-0.25)^2007*4^2008
=(1*3-1/5)*25*3+1+(-0.25*4)^2007*4
=25-15+1-4
=7

Minus one minus one-half of the negative cubic power, plus one, plus the absolute value of [(three times the root sign three minus eight times the sin60 °)]

-1-（1/2）（-³）+1+I3√3-8sin60°I
=-1-8+1+√3
=-8+√3

The real number a satisfies | 2011-a|+ a−2 012=3 a3 , find the value of a-20112

∵ a-2012 ≥ 0, that is, a ≥ 2012,
∴2011-a＜0，
∴|2011-a|+
a−2 012=a-2011+
a−2012=3 a3
=a，
Namely
a−2012=2011，
∴a-2012=20112，
Then a-20112 = 2012

If ABC is a rational number and satisfies the equation a + B radical 2 + C radical 3 = 2-radical 2 + 3 radical 3, calculate the value of (A-C) to the power of 2010 + the power of 2011 of B

Can you put a bracket or something?
A+B√2+c√3=2-√2+3*√3
According to the meaning of the title, a = 2, B = - 1, C = 3
The 2010 power of (A-C) = (2-3) = 1
B's 2011 power = - 1's 2011 power = - 1
So the final result is zero