Can the recursive equation 14.7 × 0.56-5.7 × 0.56 + 5.6 be calculated skillfully? If so, please list the formula

Can the recursive equation 14.7 × 0.56-5.7 × 0.56 + 5.6 be calculated skillfully? If so, please list the formula

It's OK
simple form
=14.7×0.56-5.7×0.56+10×0.56
=（14.7-5.7+10）×0.56
=19×0.56
=10.64

If a = 7 ^ 8, B = 8 ^ 7, please use the formula containing a and B to express the value of 56 ^ 56,

56^56
=7^56×8^56
=(7^8)^7×(8^7)^8
=a^7×b^8

What numbers should be filled in brackets 6,21,52105, ()?

It can be factorized to get
2x3 3x7 4x13 5x21
Each item is an arithmetic sequence, and the difference is 1
The difference of the second number is 4.68, which is also the arithmetic sequence of difference 2
So the next item should be 6x31 = 186

90, 81.73, (), 61, (), 52, () 44 what numbers should be filled in brackets

90,81.73,(66),61,(56),52,(49)44

61 52 63 94 46 (), what number should be filled in brackets? How did it come out?

4 * 4 = 16 is 61;
5 * 5 = 25 is 52;
6 * 6 = 36, 63;
7 * 7 = 49, 94;
8 * 8 = 64 is 46;
9 * 9 = 81, 18
So the latter one should be 18

3.141 times () is a number, then 1 / 3 () divided by 6 / 7 is 3.141, and ()

Let this number be a, then 3.141 * a * 1 / 3 * a * 7 / 6 = 3.141
The solution is: a = ± √ 14 / 6

A is a number greater than 0. If 3 / 4 divided by a is less than 3 / 4 multiplied by a, then a () 1

You can count it
Take the idiosyncrasy method
Take one that meets the requirements

795 times () divided by () = 1, 4. 587 divided by () multiplied by () = 1

795 times (1) divided by (8.795) = 1
4.587 divided by (4.587) times (1) = 1

Put the seven numbers 0, 1, 2, 3, 4, 5 and 6 in brackets to make the equation hold () x () = () = () divided by ()

(3) X (4) = (1) (2) = (60) divided by (5)

What are the rules of 9,1,4,3,40, () and what numbers should be filled in brackets

Civil service examination questions: 9, 1, 4, 3, 40, (), how many should be in brackets?
A:81 B:80 C:121 D:120
Choose D
Divide each one by three and see the remainder
The remainder is 0;
The remainder is 1;
The remainder is 1;
The remainder is 0;
The remainder is 1;
If there are two numbers with the remainder of 1, then there is a multiple before subtracting 1, that is, n = (4-1) △ 1 = 3; then there is (x-1) △ 40 = 3 (x is the sixth number)
That is: x = 40 × 3 + 1 = 121