B
1
K
Khách
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LQ
0
8 tháng 4 2018
Ta có: 1/3+1/6+1/10+...+2/x*(x+1)
=2/6+2/12+2/20+...+2/x*(x+1)
=2/2*3+2/3*4+2/4*5+...+2/x*(x+1)
=2*(1/2*3+1/3*4+1/4*5+...+1/x*(x+1))
=2*(1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1)
=2*(1/2-1/x+1)=2000/2002
=>1/2-1/x+1=2000/2002:2
=>1/2-1/x+1=500/1001
=>1/x+1=1/2-500/1001
=>1/x+1=1/2002
=>x+1=2002
=>x=2002-1
=>x=2001 thuộc N
Vậy x=2001
*Mình ko biết ấn dấu phân số với dấu nhân ở đâu, bạn thông cảm nhé!
B
1
18 tháng 5 2016
Đặt A=1/3+1/6+1/10+...+1/(x(x+1))
=> A=2/6+2/12+2/20+...+2/2(x(x+1))
=> 1/2A=1/(2.3)+1/(3.4)+...+1/2(x(x+1))
=> 1/2A=1/2-1/3+1/3-1/4+...+1/x-1/(x+1)
=> 1/2A=1/2-1/(x+1). Vì A=2000/2002=1000/1001=> 1/2A=500/1001
=> 1/2-1/(x+1)=500/1001
=> 1/(x+1)=1001/2002-1000/2002
=> 1/(x+1)=1/2002
=> x+1=2002
=> x=2001
PT
Phạm Trần Hoàng Anh
CTVHS
7 tháng 7 2020
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}\)=\(\frac{2000}{2002}\)
2.(\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\))=\(\frac{2000}{2002}\)
2.\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2000}{2002}\)
2.(\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)) = \(\frac{2000}{2002}\)
2.\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1000}{2002}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}\)
\(\frac{1}{x+1}=\frac{1}{2002}\)
2002.1 = (x+1).1
2002 = x+1
x=2001 (T/M)
MP
7 tháng 7 2020
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
\(\Rightarrow\) \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
\(\Rightarrow\) \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2000}{2002}\)
\(\Rightarrow\) \(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\Rightarrow\) \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\Rightarrow\) \(\frac{1}{2}-\frac{1}{x+1}=\frac{500}{1001}\)
\(\Rightarrow\) \(\frac{1}{x+1}=\frac{1}{2002}\)
\(\Rightarrow\) \(x+1=2002\) \(\Rightarrow\) \(x=2001\)
TM
2
HT
25 tháng 7 2015
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=\frac{2000}{2002}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{2000}{2002}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2000}{2002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{2002}:2=\frac{1000}{2002}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{1000}{2002}=\frac{1}{2002}\)
=> x + 1 = 2002
=> x = 2002 - 1
=> x = 2001
H
2
T
4 tháng 8 2018
(x+1)(x+2)+(x+3)+...+(x+100)=5750
(x+x+...+x)(1+2+3+...+100)=5750
100x + 5050 =5750
100x =700
x =7
TT
4 tháng 8 2018
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750
( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750
x . 100 + ( 100 + 1 ). 100 : 2 =5750
x .100 + 5050 = 5750
x .100 = 5750 - 5050
x . 100 = 700
x = 700 : 100
x = 7
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
12 tháng 5 2016
Đặt vế trái là A ta có:
\(\frac{A}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)
\(\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow A=\frac{x-1}{x+1}\)
\(\Rightarrow\frac{x-1}{x+1}=\frac{2007}{2009}\Leftrightarrow x=2003\)
29 tháng 11 2022
\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow...
NL
2
XO
14 tháng 7 2021
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2021}\)
<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4042}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{2042}\)
<=> \(\frac{1}{x+1}=\frac{1}{2021}\)
<=> x + 1 = 2021
<=> x = 2020
16 tháng 7 2021
Có phải là bình 6a3 học trường THCS Nguyễn Trãi đúng không
Ta có: \(A=\frac{1}{3}+\frac{1}{6}+......+\frac{2}{x.\left(x+1\right)}=\frac{2000}{2002}\)
\(A=\frac{1}{6}+\frac{1}{12}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{2002}.\frac{1}{2}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x.\left(x+1\right)}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{2}-\frac{1}{x+1}=\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2}-\frac{2000}{4004}\)
\(A=\frac{1}{x+1}=\frac{1}{2002}\)
\(x+1=2002\)
nên \(x=2002-1=2001\)
Vậy x = 2001