$50\cdot \frac{2}{5}$=20 ( students )

Number of male students is :

     50 - 20 = 30 ( students )

                                               Answer : 30 students

Number of terms is :

  ( 1000 - 1 ) : 2 + 1 = 500,5 

The sum is :

  ( 1000 + 1 ) x 500,5 : 2 = 250500,25

Number of female students is : 

   40 x 2525= 16 ( student )

 Number of  male students is :

    40 - 16 = 24 ( student )

         Answer : 24 student

 (x + 1 ) . 2000 = 4000

   x + 1 = 4000 : 2000

   x + 1 = 2 

   x = 2 - 1 

  x = 1 

We have : (x + 2)(x + 2) = 24(x + 2)(x + 2) = 24

⇒x² + 4x + 4 =24

=> x² + 4x = 20

=> x(x + 4) = 20

=> x ,x + 4 thuộc Ư(20) = {1;2;4;5;10;20}

 When x = 1 thì x + 4 = 20 => x = 16 

Not yes giá trị thoả mãn

Number of female students is : 

40 x $\frac{2}{5}$= 16 ( student )

 Number of  male students is :

    40 - 16 = 24 ( student )

         Answer : 24 student

You have:

(x+1).(1+1999)=4000

(x+1).2000=4000

x+1=4000:2000

x+1=2

x=2-1

x=1

You have:

(x+1).(1+1999)=4000

(x+1).2000=4000

x+1=4000:2000

x+1=2

x=2-1

x=1

$\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}$

$\Rightarrow \frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1$

$\Rightarrow \frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}$

a) With 2x + 3 = x + 2

=> 2x - x = 2 - 3

=> x = -1

With 2x + 3 = -(x + 2)

=> 2x + 3 = -x - 2

=> 2x + x = -2 - 3

=> 3x = -5

=> x = −53

So x = -1 and x = −53

b) We have : 

A = |x - 2006| + |2007 - x| ≥ |x - 2006 + 2007 - x| = |1| = 1

⇔{|x−2006|≤0|2007−x|≤0⇒{x=2006x=2007

So when x = 2006 ; 2007 then value of A smallest 

After 20 people arrive, the unit has all:
                 50 + 20 = 70 (people)
70 people are 50 times as many:
                 70: 50 = 7/5 (times)
70 people eat that rice in the number of days are:
                10 x 7/5 = 14 (day) (corresponding to 14 rice grades)
The additional rice capacity is:
                 14 - 10 = 4 (productivity)
                    Answer: 4 rates

After 20 people arrive, the unit has all:
                 50 + 20 = 70 (people)
70 people are 50 times as many:
                 70: 50 = 7/5 (times)
70 people eat that rice in the number of days are:
                10 x 7/5 = 14 (day) (corresponding to 14 rice grades)
The additional rice capacity is:
                 14 - 10 = 4 (productivity)
                    Answer: 4 rates

We have : ab = 8(a + b) 

=> 10a + b = 8a + 8b

=> 10a - 8a = 8b - b

=> 2a = 7b

=> a = 7 ; b = 2

So ab is 72

We have : x < -0.8 ⇒ x + 0,8 < 0 ⇒|x + 0.8| = -(x + 0.8)

                 x < -0.8 ⇒ x < 25 ⇒ |x - 25| = 25 - x

Candle A = -(x + 0.8) - (25 - x) + 1.9

                = -x + 0,8 - 25 + x + 1,9

                = -x + x - 0,8 - 25 + 1,9

                = -25,8 + 1,9

            A = -23,8

     A = -x - 0.8 - 25 + x + 1.9

     A = -23.9 

Chạy tôi EDM Sang copy my post you review time go

We find that the sum of the digits of all the years before 1899 is less than 27, so the lucky year of those born before 1899 is before 1926. So the lucky year is after 1926. That person must be born from 1900 onwards.

Assume the year of birth is 19ab = 1900 + 10 a + b, the sum of the digits is 1 + 9 + a + b = a + b +10.

So the lucky year is: 19ab + a + b +10 = 1910 + 11a + 2b.

Similarly, if another person was born in 19cd, there was a lucky year of 1910 + 11c + 2d

For two years of equal luck, 1910 + 11a + 2b = 1910 + 11c + 2a or 11 (a - c) = 2 (d - b).

If
D-b ≠ 0, so that the equality above satisfies (b - d) must be a multiple of 11, but this is absurd because b, d are digits. So d - b = 0, inference a - c = 0

Then a = c; B = d. So they gave birth the same year.

Then year must find the lucky year of a person born in 19xy and a person born in 2abc.

Numbers satisfying the problem condition are 5-digit numbers abcde with 0 <a <b <c <d <e. We see that these numbers satisfy the following two properties:

1) The numbers a, b, c, d are different (the digits are different from 0)

2) The numbers are going up from left to right.

To count the numbers satisfying the two properties simultaneously, we perform the following two steps:

Step 1: Count the 5 digit number that satisfies property 1 (different digits) that temporarily omitted property 2.

   We have:

      - There are 9 ways to choose a number (from 1 to 9)

      - After selecting a, there are 8 ways to choose the number b (from 1 to 9 but remove the selected digit).

      - After selecting a, b, there are 7 ways to choose the number c (from 1 to 9 but remove the selected number a, b)

      - After selecting a, b, c, there are 6 ways to choose d (from 1 to 9 but remove the selected a, b, c)

      - After selecting a, b, c, d, there are 5 ways to choose e (from 1 to 9 but remove the selected a, b, c, d)

   So there are all: 9 x 8 x 7 x 6 x 5 = 15120 numbers have 5 different digits and different digits 0.

Step 2: Calculate the number of numbers that satisfy the second characteristic (ie the numbers are going up from left to right).

   We have commented: For any set of five digits [eg (1, 2, 3, 4, 5)] we can write 5 x 4 x 3 x 2 x 1 = 120 different numbers from five sets This number (similar argument above).

   Of these 120 numbers, only one number satisfies the condition that the digits are going up from left to right.

   Therefore the number of numbers satisfying the above property 2 is equal to 1/120 of the numbers satisfying condition 1.

  And we have: the number of numbers satisfying simultaneously two properties 1 and 2 are: 15120: 120 = 126 numbers.

We have x + x2 + x3 + ..... + x100

At x = 1 then x + x2 + x3 + ..... + x100

= 1 + 12 + 13 + ...... + 1100 

= 1 + 1 + 1 + ..... + 1

= 1 x 100

= 100 

We have: abcdeg = abc000 + deg
                            = abc. 100000 + deg
                            = abc. 99999 + (abc + deg)
Which  99999 is divisible by 37 so abc. and abc + deg is divisible by 37

Candle 99999 is divisible by 37
So the distribution is divided by 37 (ddcm)

Put : A = 1.3 + 3.5 + 5.7 + ...... + 99.101

=> 4A = 1.3.(5 - 1) + 3.5.(7 - 3) + ...... + 99.101.(103 - 99)

=> 4A = 1.3.5 - 1.3.5 + 3.5.7 - 3.5.7 + ...... + 99.101.103

=> 4A = 99.101.103

=> A = 99.101.103/4

=> A = ..................