The 2009 power of (Radix 5 + 2) x the 2010 th power of (Radix 5-2) =

The 2009 power of (Radix 5 + 2) x the 2010 th power of (Radix 5-2) =

√5+2=(√5+2)(√5-2)/(√5-2)=1/(√5-2)
(√5+2)^2009×(√5-2)^2010
=[1/(√5-2)]^2009×(√5-2)^2010
=(√5-2)^2010/(√5-2)^2009
=√5-2

Root 2 plus root 3's 2009 power X root 2 minus root 3's 2010 power gives a result Don't give me a result just because I can't. please be responsible for me

(√2+√3)^2009*(√2-√3)^2010
=[(√2+√3)(√2-√3)]^2009*(√2-√3)
=(-1)^2009*(√2-√3)
=√3-√2

(root 3 + 2) 2009 power * (root 3-2) 2010 power This problem should be calculated first (power) or square root, then (multiplication and division), and then (addition and subtraction) ∵ (√ 3 + 2) (√ 3-2) = 3-4 = - 1 ∴(√3+2)^2009×(√3-2)^2010=(√3+2）^2009×(√3-2)^2009×(√3-2) =[(√3+2)(√3-2)]^2009×(√3-2) =-1×(√3-2) =2-√3 [knowledge points] (a + b) (a-b) = a? - B ∵ (a + b) (a-b) = a (a-b) + B (a-b) [multiplicative distributive law] =a²-ab+ba-b² =Why [(√ 3 + 2) (√ 3-2)] ^ 2009 × (√ 3-2) don't you calculate that for this one

I just want to know how educated the people I help are? My time is also time and I don't want to waste it on boring problems

The 2009 power of (7-4 Radix 3) is equal to the 2010 power of (- 7-4 Radix 3)

The original formula = (- 4 √ 3 + 7) 2009 power * (- 4 √ 3-7) 2009 power * (- 4 √ 3-7) 2009 power * (- 4 √ 3-7) = [(- 4 √ 3 + 7) (- 4 √ 3-7)] 2009 power * (- 4 √ 3-7) 2009 power * (- 4 √ 3-7) = 2009 power of 2009 power of (48-49), 2009 power of (48-49), 2009 power of (48-49), 2009 power of (- 1), 2009 power of (- 1), 2009 power of (- 1), 2009 power of (- 1), is (- 1) (- 1)) (- 1)) (- 1) (- 1) (- 1)) (- 1) (- 1)) (- 1) (- 1) (- 1) and

The 2013 power of (root 3 + root 2) is the 2014 power of (root 2-root 3)

simple form
=(√3+√2)^2013x(√3-√2)^2014
=(√3+√2)^2013x(√3-√2)^2013x(√3-√2)
=[(√3+√2)x(√3-√2)]^2013x(√3-√2)
=1^2013x(√3-√2)
=√3-√2

The 2013 power of (7-5 Radix 2) is multiplied by the 2014 power of (- 7-5 Radix 2)

simple form
=(7-5√2)^2013x(7+5√2)^2014
=(7-5√2)^2013x(7+5√2)^2013x(7+5√2)
=[(7-5√2)x(7+5√2)]^2013x(7+5√2)
=(49-50)^2013x(7+5√2)
=(-1)^2013x(7+5√2)
=-1x(7+5√2)
=-7-5√2

The negative cubic power of (√ 1 / 2) + the 2013 power of (√ 2 + 1) and the 2014 power of (root 2 + 1) - 1 It's the 2013 power of (√ 2-1). Sorry for the wrong number

The negative cubic power of (√ 1 / 2) + the 2013 power of (√ 2-1) and the 2014 power of (√ 2 + 1) - 1
=2 √ 2 + [(√ 2-1) × (√ 2 + 1)] to the power of 2013 × (√ 2 + 1) - 1
=The 2013 power of 2 √ 2 + [2-1] × (√ 2 + 1) - 1
=2√2+√2+1-1
=3√2

The 2013 power of (2 times the root number 2-3) is the 2014 power of (2 times the root number 2 + 3)

The 2013 power of (2 times the root number 2-3) is the 2014 power of (2 times the root number 2 + 3)
=The 2013 power of [(2 times root 2-3) × (2 times root 2 + 3)] is × (2 times root 2 + 3)
=The 2013 power of (8-9) × (2 times root 2 + 3)
=The 2013 power of (- 1) × (2 times the root 2 + 3)
=-(2 times root 2 + 3)
=-3-2√2

The 2014 power of (Radix 2-1 / Radix 2 + 1) is multiplied by the 2013 power of (2 times of Radix 2 + 3)

The 2014 power of (Radix 2-1 / Radix 2 + 1) is multiplied by the 2013 power of (2 times of Radix 2 + 3)
=The 2014 power of (3-2 Radix 2) is multiplied by the 2013 power of (2 times of root 2 + 3)
=The 2013 power of [(3-2 root number 2) × (2 times root number 2 + 3)] is × (3-2 root number 2)
=The 2013 power of [9-8] is × (3-2 times root number 2 + 3)
=2013 power of 1 × (3-2 root sign 2)
=3-2 times root 2

Given that a and B are rational numbers and a + radical 2B = (1-radical 2), what is the b-th power of a?

A + radical 2B = (1-radical 2) ^ 2 = 3-2 radical 2
Therefore:
a=3,b=-2
Therefore:
A^b
=3^(-2)
=1/9