After this period of review, my spelling is greatly reduced

After this period of review, my spelling is greatly reduced

After this period of review, my spelling is greatly reduced
Typos can't be reduced, they can only be reduced

The teacher said that my reading ability has increased and my composition has also improved

The teacher said that my reading ability has improved and my composition has also improved

The 55th power of 2, the 44th power of 3, and the 33rd power of 4 are relatively large. The one in the middle is not easy to do

Because: the 55th power of 2 = the 11th power of 32
44th power of 3 = 11th power of 81
33 power of 4 = 11 power of 64
So: the 44th power of 3 > the 33rd power of 4 > the 55th power of 2

New year's day in 2008 is Tuesday, and the 2008 Beijing Olympic Games will open on August 8. What day is that?
Write out the calculation process

30 + 29 + 31 + 30 + 31 + 30 + 31 + 8 = 220 (days)
220 / 7 = 31 ^ 3 (days)
Friday

A class of 46 students to go boating, a total of 10 boats, 6 people in the big boat, 4 people in the small boat, all full?

If there are x large boats, there are 10-x small boats
6X+4(10-X)=46
6X-4X=46-40
X=3
10-3=7
There are three big boats and seven small boats

The square of (- 99 and a half)

Mathematics (derivation of sequence formula)
Find 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + n^2
What is the formula and how to deduce it

1^2+2^2+3^2+…… +n^2=n(n+1)(2n+1)/6
Using the formula of cubic difference
n^3-(n-1)^3=1*[n^2+(n-1)^2+n(n-1)]
=n^2+(n-1)^2+n^2-n
=2*n^2+(n-1)^2-n
2^3-1^3=2*2^2+1^2-2
3^3-2^3=2*3^2+2^2-3
4^3-3^3=2*4^2+3^2-4
.
n^3-(n-1)^3=2*n^2+(n-1)^2-n
Add all the equations
n^3-1^3=2*(2^2+3^2+...+n^2)+[1^2+2^2+...+(n-1)^2]-(2+3+4+...+n)
n^3-1=2*(1^2+2^2+3^2+...+n^2)-2+[1^2+2^2+...+(n-1)^2+n^2]-n^2-(2+3+4+...+n)
n^3-1=3*(1^2+2^2+3^2+...+n^2)-2-n^2-(1+2+3+...+n)+1
n^3-1=3(1^2+2^2+...+n^2)-1-n^2-n(n+1)/2
3(1^2+2^2+...+n^2)=n^3+n^2+n(n+1)/2=(n/2)(2n^2+2n+n+1)
=(n/2)(n+1)(2n+1)
1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
1^3+2^3+3^3+…… +n^3=[n(n+1)/2]^2
(n+1)^4-n^4=[(n+1)^2+n^2][(n+1)^2-n^2]
=(2n^2+2n+1)(2n+1)
=4n^3+6n^2+4n+1
2^4-1^4=4*1^3+6*1^2+4*1+1
3^4-2^4=4*2^3+6*2^2+4*2+1
4^4-3^4=4*3^3+6*3^2+4*3+1
.
(n+1)^4-n^4=4*n^3+6*n^2+4*n+1
What is the sum of all forms
(n+1)^4-1=4*(1^3+2^3+3^3...+n^3)+6*(1^2+2^2+...+n^2)+4*(1+2+3+...+n)+n
4*(1^3+2^3+3^3+...+n^3)=(n+1)^4-1+6*[n(n+1)(2n+1)/6]+4*[(1+n)n/2]+n
=[n(n+1)]^2
1^3+2^3+...+n^3=[n(n+1)/2]^2

How much is one third plus three fifths multiplied by 15

If the values of 3x-1 and 4x + 8 are equal, then x =?

If a divided by 3 / 4 equals B divided by 2 / 3, which of a and B is the largest

a/(3/4)=b/(2/3)
=> a*(4/3)=b*(3/2)
=> a/b=(3/2)*(3/4)
=> a/b = 9/8
If a and B are both 0, they are the same size
A and B are positive numbers, a big, negative numbers, B big