1-8, fill in the box with 8 numbers, so that the sum of each side is equal to 15

1-8, fill in the box with 8 numbers, so that the sum of each side is equal to 15

8 1 6
4 2
3 5 7

Fill 0 to 9 in the box □ + □ = □ + □ = □ + □ = □ + □ = □ + □ = □ + □, so that the equation is true and the numbers are not repeated

0+9=1+8=2+7=3+6=4+5

() () - () = () () = () × () () 1 - nine numbers are filled in the first nine boxes to make the equation tenable and can't be repeated

17×4=68=93-25

Fill in 1 to 9 numbers - + - = - so that the three equations can't be repeated
Yes () + () = (), () can only fill in one number

According to you said no solution, similar I found you a post
Because if you add up these three formulas, it's equivalent to the sum of six numbers and equal to the sum of three numbers
Then it can only be even = even or odd = odd, that is, the sum of all numbers is even
But the sum of 1 to 9 is 45, which is odd, so the above equation cannot be true

If I is an imaginary unit, then 1 + i1-i=______ ．

1 + i1-i = (1 + I) (1 + I) (1-I) (1 + I) = 2I2 = I, so the answer is: I

I is an imaginary unit, I / (1 + I) =?

i/(1+i)
=i(1-i)/[(1+i)(1-i)]
=(-i^2+i)/(1-i^2)
=(1+i)/(1+1)
=(1/2)+(1/2)i
i^2=-1

If I is an imaginary unit, then (1 + I) I = what

i-1

Why is the absolute value of imaginary unit I 1

In the complex number set, the imaginary number unit is I, which means I & # 178; = - 1. The so-called absolute value of the imaginary number unit is actually the feeling of the complex number I [for the complex number Z = a + bi, its module is | Z | = √ (A & # 178; + B & # 178;)]. Obviously, the module of the imaginary number unit is 1

How many zeros are there at the end of the result calculated by the formula 50 * 53 * 56 * ·· * 110

How many zeros are required at the end of this formula? We need to calculate how many pairs of 2 and 5 there are after decomposing the prime factor
After decomposition, there are 20 in 2 and 6 in 5
So there are six zeros at the end of this formula

(4x-1) divided by (X-2) (x-1) = m divided by (X-2) plus x divided by (x-1), then M equals

M = 1-x (x ≠ 1 and - 2)
Calculation after general division