rational numbers
Cristiano Ronaldo
19/03/2017 at 11:36
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FA KAKALOTS 03/02/2018 at 12:43
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
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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
\(\Rightarrow A=\dfrac{3^{64}-1}{2}\)
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Phan Huy Toàn 15/08/2017 at 08:36
bạn không nên đăng những câu hỏi linh tinh như 1+1,... trên diễn đàn
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FA KAKALOTS 03/02/2018 at 12:43
(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
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Phan Duy Truong ❀◕ ‿ ◕❀ 26/03/2017 at 08:46
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{1999^2}\right)\left(1-\dfrac{1}{2000^2}\right)=\dfrac{1\times3}{2\times2}\times\dfrac{2\times4}{3\times3}\times\dfrac{3\times5}{4\times4}\times...\times\dfrac{1999\times2001}{2000\times2000}\)
\(=\dfrac{\left(1\times2\times3\times...\times1999\right)\left(3\times4\times5\times...\times2001\right)}{\left(2\times3\times4\times...\times2000\right)\left(2\times3\times4\times...\times2000\right)}\)
\(=\dfrac{1\times2001}{2000\times2}=\dfrac{2001}{4000}\)
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FA KAKALOTS 03/02/2018 at 12:43
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
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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 :
\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+4+...+50}\)\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{50.51}\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{50.51}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{51}\right)=1-\dfrac{2}{51}=\dfrac{49}{51}\)
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FA KAKALOTS 03/02/2018 at 12:44
[1(1−12)(1−13)(1−14)(1−15)...(1−12000)]×2000
=12.23.34.45....19992000.2000=1
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\(\left[1\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2000}\right)\right]\times2000\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}....\dfrac{1999}{2000}.2000=1\)
Cristiano Ronaldo
19/03/2017 at 10:59
Cristiano Ronaldo
19/03/2017 at 10:56
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FA KAKALOTS 03/02/2018 at 12:44
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
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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 :
\(\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}\right)+\left(\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}\right)+...+\left(\dfrac{1}{100}+\dfrac{2}{100}+\dfrac{3}{100}+...+\dfrac{99}{100}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{99}{2}\)
\(=\dfrac{\dfrac{99.100}{2}}{2}=\dfrac{99.100}{4}=99.25=2475\)
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Nguyễn Kim Ngưu 19/03/2017 at 11:00
\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n.\left(n+1\right)}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{n}-\dfrac{1}{n-1}\)
\(A=1-\dfrac{1}{n+1}=\dfrac{n}{n+1}=\dfrac{1999}{2000}\Rightarrow n.2000=\left(n+1\right).1999\)
\(\Leftrightarrow1999n+n=1999n+1999\Rightarrow n=1999\)
Cristiano Ronaldo selected this answer. -
Nguyệt Nguyệt 19/03/2017 at 12:14
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. -
FA KAKALOTS 03/02/2018 at 12:44
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.
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caubevodanh3 19/03/2017 at 11:31
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
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FA KAKALOTS 03/02/2018 at 12:45
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
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FA KAKALOTS 03/02/2018 at 12:45
2000(1+12)(1+13)(1+14)...(1+12000)
=2000.32.43.54....20012000=2000.20012
= 1000.2001 = 2001000
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\(2000\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(1+\dfrac{1}{2000}\right)\)
\(=2000.\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}....\dfrac{2001}{2000}=2000.\dfrac{2001}{2}\)= 1000.2001 = 2001000
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FA KAKALOTS 03/02/2018 at 12:45
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 -
Nguyệt Nguyệt 19/03/2017 at 18:18
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 = \(\dfrac{23}{70}\)
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FA KAKALOTS 03/02/2018 at 12:46
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)
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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)
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Dung Trần Thùy 18/03/2017 at 18:54
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 :>
steve jobs selected this answer. -
Lê Anh Tú 25/03/2017 at 22:18
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)
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FA KAKALOTS 03/02/2018 at 12:46
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 :>
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Trần Quang Linh 17/03/2017 at 15:53
A=3+32+33+...+32000
3A=32+33+34+...+32001
2A=32001-3
A=\(\dfrac{3^{2001}-3}{2}\)
steve jobs selected this answer. -
Love people Name Jiang 17/03/2017 at 18:41
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
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FA KAKALOTS 03/02/2018 at 12:46
Ta có : 3 + 32 + 33 + ...... + 32000
=> 3A = 32 + 33 + 34 + ...... + 32001
=> 3A - A = 32001 - 3
=> 2A = 32001 - 3
=> A = 32001−32
Good
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Trần Quang Linh 17/03/2017 at 15:59
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=\(\dfrac{127}{128}\)
steve jobs selected this answer. -
Love people Name Jiang 17/03/2017 at 18:43
For : \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+.....+\dfrac{1}{128}\)
\(\Rightarrow2A-A=1-\dfrac{1}{128}\)
\(\Rightarrow A=\dfrac{127}{128}\)
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FA KAKALOTS 03/02/2018 at 12:46
For : A=12+14+18+132+164+1128
⇒2A=1+12+14+.....+1128
⇒2A−A=1−1128
⇒A=127128
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Trần Quang Linh 17/03/2017 at 16:02
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=\(\dfrac{999}{1000}\)
steve jobs selected this answer. -
Love people Name Jiang 17/03/2017 at 18:37
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{999999.1000}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.......+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(=1-\dfrac{1}{1000}\)
\(=\dfrac{999}{1000}\)
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Đại Việt 17/03/2017 at 21:09
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=\(1-\dfrac{1}{1000}=\dfrac{1000}{1000}-\dfrac{1}{1000}=\dfrac{999}{1000}\)
steve jobs
17/03/2017 at 14:09
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FA KAKALOTS 03/02/2018 at 12:47
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
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Trần Quang Linh 17/03/2017 at 16:21
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 \(\dfrac{1}{277x257}\)
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\(\dfrac{2000.2001}{2002.2003}< \dfrac{2000.2002}{2002.2003}< \dfrac{2000.2002}{2001.2003}< \dfrac{2000.2003}{2001.2003}< \dfrac{2000.2003}{2001.2002}\)
=> C < B < A
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FA KAKALOTS 03/02/2018 at 12:48
2000.20012002.2003<2000.20022002.2003<2000.20022001.2003<2000.20032001.2003<2000.20032001.2002
=> C < B < A