The 2002 power of (Radix 2-radix 3) is multiplied by the 2003 power of (Radix 2 + Radix 3)

The 2002 power of (Radix 2-radix 3) is multiplied by the 2003 power of (Radix 2 + Radix 3)

Solution; (root 2-root 3) the power of 2002 times (root 2 + root 3) = (root 2-root 3) ^ 2002 * (root 2 + root 3) ^ 2002 * (root 2 + root 3)
=[(root 2-root 3) (root 2 + root 3)] ^ 2003 * (root 2 + root 3)
=(2-3) ^ 2003 (root 2 + root 3)
=-1 (root 2 + root 3)
=-Radical 2-radical 3

The 2002 power of (Radix 6-radix 5) is multiplied by the 2001 power of (Radix 6-radix 5)

The results were: root 6-root 5
The 2002 power of (Radix 6-radix 5) is multiplied by the 2001 power of (Radix 6-radix 5) =
(Radix 6-radix 5) is multiplied by the 2001 power of (Radix 6-radix 5) times (Radix 6-radix 5) =
1 × (root 6-root 5)

What is the 2004 power of (Radix 5 minus 2) multiplied by the 2003 power of (Radix 5 + 2)?

The 2004 power of (Radix 5 minus 2) is multiplied by the 2003 power of (Radix 5 + 2)
Is equal to the 2003 power of (root 5 minus 2) times (root 5 plus 2) 2003 times (root 5 minus 2)
Is equal to the 2003 power of ((root 5 minus 2) times (root 5 plus 2)) (root 5 minus 2)
(root 5 minus 2) times (root 5 + 2) = 5-4 = 1
So the 2004 power of (Radix 5 minus 2) is multiplied by the 2003 power of (Radix 5 + 2)
The 2003 power of 1 (root 5 minus 2)
Equal to (root 5 minus 2)

How much is the 2003 power of (Radix 5 + 2) multiplied by the 2004 power of (Radix 5-2)

(Radix 5 + 2) * (Radix 5-2) = 5-4 = 1
So (radical 5 + 2) ^ 2003 × (Radix 5-2) ^ 2004
=(Radix 5 + 2) ^ 2003 × (Radix 5 + 2) ^ (- 2004)
=(Radix 5 + 2) ^ (2003-2004)
=(radical 5 + 2) ^ (- 1)
=Radical 5-2

The 2003 power of (- 2-radical 3) x (2-radical 3) to the 2004 power=

The 2003 power of (- 2-radical 3) * (2-radical 3) to the 2004 power
=The 2003 power of (- 2-radical 3) * (2-radical 3) * (2-radical 3)
=[(- 2-radical 3) (2-radical 3)] ^ 2003 * [2-3 ^ (1 / 2)] = (- 1) ^ 2003 * [2-3 ^ (1 / 2)] = - [2-3 ^ (1 / 2)] = 3 ^ (1 / 2) - 2

Calculate the 2001 power of {(2) minus (3)} times {2002 power of (2) plus (3)] I counted - 5, right?

(√2-√3)^2001×(√2+√3)^2002
=[(√2-√3)(√2+√3)]^2001×(√2+√3)
=(-1)^2001×(√2+√3)
=-1×(√2+√3)
= -√2-√3.

The third power of a root 8a-a square root one half a + 3 root 2A

a√(8a)=2a√(2a)
a√(0.5a)=0.5a√(2a)
3√(2a*a*a)=3a√(2a)
So a √ (8a) - a √ (0.5A) + 3 √ (2a * a * a) = 4.5a √ (2a)

(negative half) - power 1 - root sign 12 + (1 - radical 2) power 0 - | root sign 3 - root sign 2|

(negative half) - power 1 - root sign 12 + (1 - radical 2) power 0 - | root sign 3 - root sign 2|
=(negative 2) power 1 - root sign (4x3) + 1 - (radical 3 - root 2)
=-2-2 root sign (3) + 1 - root sign 3 + root sign 2
=-1-3 root number 3 + root number 2

Calculation: the negative first power of | - root 3 | + (one half) - 3 / root 3

=√3+2-√3=2

One half of two equals plus or minus root two or root two? Why? As the title. Tell me why

The second power of two is the root two
The square root of two is positive and negative root two
The arithmetic square root of two is root two