rational numbers

Let A = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)

=> 2A = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)

           = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)

           = (34 - 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)

           = (38 - 1)(38 + 1)(316 + 1)(332 + 1)

           = (316 - 1)(316 + 1)(332 + 1)

           = (332 - 1)(332 + 1)

           = 364 - 1

⇒A=364−12

Let A = (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

=> 2A = (3 - 1)(3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

           = (3² - 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

           = (3⁴ - 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

           = (3⁸ - 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1)

           = (3¹⁶ - 1)(3¹⁶ + 1)(3³² + 1)

           = (3³² - 1)(3³² + 1)

           = 3⁶⁴ - 1

\(\Rightarrow A=\dfrac{3^{64}-1}{2}\)

bạn không nên đăng những câu hỏi linh tinh như 1+1,... trên diễn đàn

(1−122)(1−132)(1−142)...(1−119992)(1−120002)=1×32×2×2×43×3×3×54×4×...×1999×20012000×2000

=(1×2×3×...×1999)(3×4×5×...×2001)(2×3×4×...×2000)(2×3×4×...×2000)

=1×20012000×2=20014000

\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{1999^2}\right)\left(1-\dfrac{1}{2000^2}\right)=\dfrac{1\times3}{2\times2}\times\dfrac{2\times4}{3\times3}\times\dfrac{3\times5}{4\times4}\times...\times\dfrac{1999\times2001}{2000\times2000}\)

\(=\dfrac{\left(1\times2\times3\times...\times1999\right)\left(3\times4\times5\times...\times2001\right)}{\left(2\times3\times4\times...\times2000\right)\left(2\times3\times4\times...\times2000\right)}\)

\(=\dfrac{1\times2001}{2000\times2}=\dfrac{2001}{4000}\)

Consider the following expression :

11+2+3+..+n=1n(n+1)2=2n(n+1)

So we have :

11+2+11+2+3+11+2+3+4+...+11+2+3+4+...+50

=22.3+23.4+24.5+...+250.51

=2(12.3+13.4+14.5+...+150.51)

=2(12−13+13−14+14−15+...+150−151)

=2(12−151)=1−251=4951

Consider the following expression :

\(\dfrac{1}{1+2+3+..+n}=\dfrac{1}{\dfrac{n\left(n+1\right)}{2}}=\dfrac{2}{n\left(n+1\right)}\)

So we have :

\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+50}\)\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{50.51}\)

\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{50.51}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{51}\right)=1-\dfrac{2}{51}=\dfrac{49}{51}\)

[1(1−12)(1−13)(1−14)(1−15)...(1−12000)]×2000

=12.23.34.45....19992000.2000=1

\(\left[1\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2000}\right)\right]\times2000\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{1999}{2000}.2000=1\)

Consider the following expression :

1n+1+2n+1+3n+1+...+nn+1

=n(n+1)2n+1=n2

So we have :

12+(13+23)+(14+24+34)+(15+25+35+45)+...+(1100+2100+3100+...+99100)

=12+22+32+42+...+992

=99.10022=99.1004=99.25=2475

Consider the following expression :

\(\dfrac{1}{n+1}+\dfrac{2}{n+1}+\dfrac{3}{n+1}+...+\dfrac{n}{n+1}\)

\(=\dfrac{\dfrac{n\left(n+1\right)}{2}}{n+1}=\dfrac{n}{2}\)

So we have :

\(\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+...+\left(\dfrac{1}{100}+\dfrac{2}{100}+\dfrac{3}{100}+...+\dfrac{99}{100}\right)\)

\(=\dfrac{1}{2}+\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{99}{2}\)

\(=\dfrac{\dfrac{99.100}{2}}{2}=\dfrac{99.100}{4}=99.25=2475\)

\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n.\left(n+1\right)}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{n}-\dfrac{1}{n-1}\)

\(A=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}=\dfrac{1999}{2000}\Rightarrow n.2000=\left(n+1\right).1999\)

\(\Leftrightarrow1999n+n=1999n+1999\Rightarrow n=1999\)

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.

A=12+16+112+...+1n(n+1)=11.2+12.3+...+1n.(n+1)

A=1−12+12−13+..+1n−1n−1

A=1−1n+1=nn+1=19992000⇒n.2000=(n+1).1999

⇔1999n+n=1999n+1999⇒n=1999

We have :

12+16+112+...+1n.(n+1)=19992000

= 11.2+12.3+13.4+...+1n.(n+1)=19992000
= 1−12+12−13+13−14+...+1n−1n+1=19992000
=  1−1n+1=19992000
= > 1n+1=1−19992000
<=> 1n+1=12000
=> n + 1 = 2000
n = 2000 - 1
n = 1999
Vậy n = 1999.

Put : A= 1 +2 + 2² + 2³ + ..... + 2²⁰⁰⁰

 => 2A = 2 x ( 1+ 2 + 2² + 2³ +....+ 2²⁰⁰⁰ )

     2A = 2 + 2² + 2³ + ..... + 2²⁰⁰¹

=> 2A-A= (2 + 2² + 2³ + ..... + 2²⁰⁰¹)- ( 1 +2 + 2² + 2³ + ..... + 2²⁰⁰⁰)

=> A = 2²⁰⁰¹ - 1

Put : A= 1 +2 + 22 + 23 + ..... + 22000

 => 2A = 2 x ( 1+ 2 + 22 + 23 +....+ 22000 )

     2A = 2 + 22 + 23 + ..... + 22001

=> 2A-A= (2 + 22 + 23 + ..... + 22001)- ( 1 +2 + 22 + 23 + ..... + 22000)

=> A = 22001 - 1

2000(1+12)(1+13)(1+14)...(1+12000)

=2000.32.43.54....20012000=2000.20012

= 1000.2001 = 2001000

\(2000\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(1+\dfrac{1}{2000}\right)\)

\(=2000.\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{2001}{2000}=2000.\dfrac{2001}{2}\)= 1000.2001 = 2001000

Gọi biểu thức đó là A. Ta có:
A=11.4+14.7+17.10+...+167.70
3A = 3.(11.4+14.7+17.10+...+167.70)
3A = 31.4+34.7+37.10+...+367.70
3A = 1−14+14−17+17−110+...+167−170
3A = 1−170
3A = 6970
=> A = 6970:3
A = 6970.13
A = 2370

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 :

3000.30013002.3003<3000.30023002.3003<3000.30023001.3003<3000.30033001.3003<3000.30033001.3002

(because 3000.3001 < 3000.3002 < 3000.3003 ;

               3002.3003 > 3001.3003 > 3001.3002)

Hence C < B < A

So the correct answer is (A)

We have :

\(\dfrac{3000.3001}{3002.3003}< \dfrac{3000.3002}{3002.3003}< \dfrac{3000.3002}{3001.3003}< \dfrac{3000.3003}{3001.3003}< \dfrac{3000.3003}{3001.3002}\)

(because 3000.3001 < 3000.3002 < 3000.3003 ;

               3002.3003 > 3001.3003 > 3001.3002)

Hence C < B < A

So the correct answer is (A)

Hi Steve Jobs! First; you can see : 3 . 5 = 15 ; 33 . 35 = 1155;

333 . 335 = 111555, ... It's the first step to solve ur problem.

Now, we have to prove that expression equal 333...3 . 333....5.

We have :

\(111...11111111111555...555555555\)

( 2002 1s)                      (2002 5s)

=111.....11000....0 + 555.......5 

( 2002 1s) (2002 0s) (2002 5s)

= 1111.....111 . ( 10000...000 + 5 )

    ( 2002 1s)          ( 2002 0s)

= 111....111 .  10000...00005

    ( 2002 1s)       (2001 0s )

= 1111...1111 . ( 3 . 333...33335 )

  ( 2002 1s)               (2001 3s )

= 333......3333 . 333333...3335

       ( 2002 3s)     ( 2001 3s )

The sum of these 2 numbers is 6666......68

                                                  (2002 6s)

Sorry if my English is bad :>

Hi Steve Jobs! First; you can see : 3 . 5 = 15 ; 33 . 35 = 1155;

333 . 335 = 111555, ... It's the first step to solve ur problem.

Now, we have to prove that expression equal 333...3 . 333....5.

We have :

111...11111111111555...555555555111...11111111111555...555555555

( 2002 1s)                      (2002 5s)

=111.....11000....0 + 555.......5 

( 2002 1s) (2002 0s) (2002 5s)

= 1111.....111 . ( 10000...000 + 5 )

    ( 2002 1s)          ( 2002 0s)

= 111....111 .  10000...00005

    ( 2002 1s)       (2001 0s )

= 1111...1111 . ( 3 . 333...33335 )

  ( 2002 1s)               (2001 3s )

= 333......3333 . 333333...3335

       ( 2002 3s)     ( 2001 3s )

The sum of these 2 numbers is 6666......68

                                                  (2002 6s)

Hi Steve Jobs! First; you can see : 3 . 5 = 15 ; 33 . 35 = 1155;

333 . 335 = 111555, ... It's the first step to solve ur problem.

Now, we have to prove that expression equal 333...3 . 333....5.

We have :

111...11111111111555...555555555

( 2002 1s)                      (2002 5s)

=111.....11000....0 + 555.......5 

( 2002 1s) (2002 0s) (2002 5s)

= 1111.....111 . ( 10000...000 + 5 )

    ( 2002 1s)          ( 2002 0s)

= 111....111 .  10000...00005

    ( 2002 1s)       (2001 0s )

= 1111...1111 . ( 3 . 333...33335 )

  ( 2002 1s)               (2001 3s )

= 333......3333 . 333333...3335

       ( 2002 3s)     ( 2001 3s )

The sum of these 2 numbers is 6666......68

                                                  (2002 6s)

Sorry if my English is bad :>

A=3+3²+3³+...+3²⁰⁰⁰

3A=3²+3³+3⁴+...+3²⁰⁰¹

2A=3²⁰⁰¹-3

A=\(\dfrac{3^{2001}-3}{2}\)

Ta có : 3 + 3² + 3³ + ...... + 3²⁰⁰⁰

=> 3A = 3² + 3³ + 3⁴ + ...... + 3²⁰⁰¹

=> 3A - A = 3²⁰⁰¹ - 3

=> 2A = 3²⁰⁰¹ - 3

=> A = \(\dfrac{3^{2001}-3}{2}\)

Good

Ta có : 3 + 32 + 33 + ...... + 32000

=> 3A = 32 + 33 + 34 + ...... + 32001

=> 3A - A = 32001 - 3

=> 2A = 32001 - 3

=> A = 32001−32

Good

A=\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

128A=64+32+16+8+4+2+1

128A=127

A=\(\dfrac{127}{128}\)

For : \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+.....+\dfrac{1}{128}\)

\(\Rightarrow2A-A=1-\dfrac{1}{128}\)

\(\Rightarrow A=\dfrac{127}{128}\)

For : A=12+14+18+132+164+1128

⇒2A=1+12+14+.....+1128

⇒2A−A=1−1128

⇒A=127128

A=\(\dfrac{1}{1x2}+\dfrac{1}{2x3}+...+\dfrac{1}{999x1000}\)

A=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)

A=\(1-\dfrac{1}{1000}\)

A=\(\dfrac{999}{1000}\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{999999.1000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.......+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(=1-\dfrac{1}{1000}\)

\(=\dfrac{999}{1000}\)

Call A=\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{999+1000}\)

We have A=\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{999\cdot1000}\)

\(\Rightarrow\)A=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)

\(\Rightarrow\)A=\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{999}-\dfrac{1}{999}\right)-\dfrac{1}{1000}\)\(\Rightarrow\) A=\(1-\dfrac{1}{1000}=\dfrac{1000}{1000}-\dfrac{1}{1000}=\dfrac{999}{1000}\)

It is too long so I don't want to write too much

use The nature of multiplication with addition to divide 1277

The answer is 1277x257

It is too long so I don't want to write too much

use The nature of multiplication with addition to divide \(\dfrac{1}{277}\)

The answer is \(\dfrac{1}{277x257}\)

\(\dfrac{2000.2001}{2002.2003}< \dfrac{2000.2002}{2002.2003}< \dfrac{2000.2002}{2001.2003}< \dfrac{2000.2003}{2001.2003}< \dfrac{2000.2003}{2001.2002}\)

=> C < B < A

2000.20012002.2003<2000.20022002.2003<2000.20022001.2003<2000.20032001.2003<2000.20032001.2002

=> C < B < A