(1-2 to the second power) (1-3 to the second power) (one of the second powers of 1-2010) (one of the second powers of 1-2011)

Using the square difference formula expansion = (1-1 / 2) (1 + 1 / 2) (1-1 / 3) (1 + 1 / 3) (1-1 / 4) (1 + 1 / 4) (1-1/2011)(1+1/2011)=1/2× 3/2×2/3 × 4/3×3/4× 5/4×4/5×…… ×2010/2011 ×2012/2011=1/2×2012/2011=1006/2011

Use a simple method to calculate the (3rd power of 2010 + 2nd power of 2010-2011) (3rd power of 2010-2 × 2nd power of 2010-2008)

Let a = 2010
The original formula = (2010 ^ 3 + 2010 ^ 2-2011) / (2010 ^ 3-2 * 2010 ^ 2-2008)
=(A^3+A^2-A-1)/(A^3-2A^2-A+2)
=(A+1)(A^2-1)/[A^3-A^2-(A^2+A-2)]
=(A+1)^2(A-1)/[A^2(A-1)-(A-1)(A+2)]
=(A+1)^2(A-1)/(A-1)(A^2-A-2)
=(A+1)^2(A-1)/(A-1)(A-2)(A+1)
=(A+1)/(A-2)
=(2010+1)/(2010-2)
=2011/2008
In the case of large or complex numbers or formulas, it is better to use the idea of substitution

Calculation: the 2nd power of 2010 - the 2nd power of 2009 + the 2nd power of 2008 -. + the 2nd power of 2 - the 2nd power of 1

The 2nd power of 2010 - the 2nd power of 2009 + the 2nd power of 2008 -. + the 2nd power of 2 - the 2nd power of 1
=(2010+2009)+(2008+2007)+.+(2+1)
=(2010+1)*2010/2
=2021055

It is known that: the m power of 10 = 20, the n power of 10 = the - 1 power of 5. What is the value of 9m divided by 3 to the 2n power

It should be the m power of 9. If you think it is good, use it!
Given that 10 ^ m = 20,10 ^ n = 5 ^ (- 1), find 9 ^ m / (3 ^ 2n) =?
10^m=20,10^n=5^(-1)=1/5
Divide the above two formulas to get
10^m÷10^n=20÷1/5=20×5=100
10^(m-n)=10^2
m-n=2
therefore
9^m÷3^2n
=9^m÷9^n
=9^(m-n)
=9^2
=81
In the computer, ^ "means how many times

What is the 100th power of 2 × (- 1 / 2) to the 101th power

The 100th power of 2 × (- 1 / 2) is the 101th power of (- 1 / 2)
=The 100th power of 2 × (- 1 / 2) × (- 1 / 2)
=The 100th power of [2 × (- 1 / 2)]
=(- 1) to the 100th power × (- 1 / 2)
=1×(-1/2)
=-1/2

Given the m power of a = 3 and the n power of a = 2, find the value of a to the power of 2m-3n

a^2m-3n
=a^2m÷a^3n
=(a^m)²÷(a^n)³
=3²÷2³
=9÷8
=9/8

Given the m power of a = 5 and the n power of a = 3, what is the value of 2m + 3N power of a?

a^m=5,a^n=3
a^(2m+3n)
=a^(2m)*a^(3n)
=(a^m)²*(a^n)³
=5²*3³
=675

If 2x's nth power + (m-1) x + 1 is a cubic binomial, find the square of M - the square of N?

The n-th power of 2x + (m-1) x + 1 is a cubic binomial
So n is the highest order and the term is twice. N = 2
At the same time, m-1=0, to ensure that there are only two items
So m = 1, n = 2
Square of M - square of n
=1-4=-3

It is known that in the solutions of the system of bivariate linear equations 3x + 5Y = m + 2,2x + 3Y = m, X is greater than y by 1. Find the value of M and the solution of the system of equations

3x+5y=m+2
2x+3y=m
X is 1 greater than y
So x = y + 1
Put in two equations
3y+3+5y=m+2
m=8y+1
2y+2+3y=m
m=5y+2
So 8y + 1 = 5Y + 2
y=1/3
x=y+1=4/3
m=8y+1=11/3

If n is a positive integer and the 2n power of a is equal to 5, what is the (3N power of 2a) 2 divided by the 4N power of 4A? Find the final number and write the reason

a^(2n)=5
(2a^3n)^2/(4a^4n)
=2a^6n/4a^4n
=a^2n/2
=5/2

(1-2 to the second power) (1-3 to the second power) (one of the second powers of 1-2010) (one of the second powers of 1-2011)