123476543 = 11111 x 11113

Gọi biểu thức đó là A. Ta có:
A=\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{67.70}\)
3A = \(3.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{67.70}\right)\)
3A = \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{67.70}\)
3A = \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{67}-\dfrac{1}{70}\)
3A = \(1-\dfrac{1}{70}\)
3A = \(\dfrac{69}{70}\)
=> A = \(\dfrac{69}{70}:3\)
A = \(\dfrac{69}{70}.\dfrac{1}{3}\)
A = \(\dfrac{23}{70}\)

We have :

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{1999}{2000}\)
= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{1999}{2000}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1999}{2000}\)
=  \(1-\dfrac{1}{n+1}=\dfrac{1999}{2000}\)
= > \(\dfrac{1}{n+1}=1-\dfrac{1999}{2000}\)
<=> \(\dfrac{1}{n+1}=\dfrac{1}{2000}\)
=> n + 1 = 2000
n = 2000 - 1
n = 1999
Vậy n = 1999.

3x - 2y = 5
2y + 3z = 1
x - z = 4
We have :
( 3x - 2y ) - ( 2y + 3z ) = 5 - 1
3x - 2y - 2y - 3z = 4
3x - 4y  - 3z = 4
3x - 3z - 4y = 4
3. ( x - z ) - 4y = 4
3 . 4 - 4y  = 4   ( replace x - z = 4 )
12 - 4y  = 4
       4y   = 12 - 4
      4y     = 8
        y     = 8 : 4
       y      =  2
=> 3x - 2.2 = 5    ( replace y = 2 )
      3x - 4 = 5
      3x       = 5 + 4
      3x       = 9
        x       = 9 : 3
        x       = 3
=> 3 - z = 4  ( replace x = 3 )
           z  = 3 - 4
            z = -1
    So y = 2, x = 3 and z = -1

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}\)
\(A=\dfrac{99}{100}\)

Hello. My name is Nguyet Nguyet. I am thirteen years old and i study at Vo Thi Sau secondary school.

3x - 2y = 5
2y + 3z = 1
x - z = 4
We have :
( 3x - 2y ) - ( 2y + 3z ) = 5 - 1
3x - 2y - 2y - 3z = 4
3x - 4y  - 3z = 4
3x - 3z - 4y = 4
3. ( x - z ) - 4y = 4
3 . 4 - 4y  = 4   ( replace x - z = 4 )
12 - 4y  = 4
       4y   = 12 - 4
      4y     = 8
        y     = 8 : 4
       y      =  2
=> 3x - 2.2 = 5    ( replace y = 2 )
      3x - 4 = 5
      3x       = 5 + 4
      3x       = 9
        x       = 9 : 3
        x       = 3
=> 3 - z = 4  ( replace x = 3 )
           z  = 3 - 4
            z = -1
    So y = 2, x = 3 and z = -1

a)  ( 2x + 3 )² 
= ( 2x + 3 ) . ( 2x + 3 )
= 2x² + 6x  + 6x + 9
= 2x² + 12x + 9
= x( 2x + 12 ) + 9

I consider red marble is 3x + 1, yellow marble is 3x + 2, green marble is 3x ( With x \(\in\) N )
Because 2017 = 3x + 1 should :
Color of the 2017^th is red marble.

We have :

A = \(\dfrac{1}{2016}+\dfrac{3}{2016}+\dfrac{5}{2016}+...+\dfrac{2015}{2016}\)
A = \(\dfrac{1+3+5+...+2015}{2016}\)
A = \(\dfrac{1016064}{2016}=504\)
=> Sentence c.