Tìm x
a) 1/3+1/6+1/10+...+1/x.(x+1)=-2007/2009
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NN
0
ZV
3
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...
N
1
7 tháng 5 2015
ta có: 1/3 + 1/6 + ... + 2/x(x+1) = 2/2.3 + 2/3.4 +.......2/x(x+1) = 2(1/2.3 +1/3.4 +.....+1/x(x+1)) = 2.(1/2-1/3+1/3-1/4+....+1/x-1/(x+1))= 2.(1/2-1/(x+1)) = 1-2/(x+1)
giải 1-2/(x+1) = 2007/2009 ta được x=2008
VT
1
DD
Đoàn Đức Hà
Giáo viên
15 tháng 5 2021
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}\)
\(=2\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x\left(x+1\right)}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{x+1}\right)\)
\(=1-\frac{2}{x+1}\)
Phương trình ban đầu tương đương với:
\(1-\frac{2}{x+1}=\frac{2007}{2009}\)
\(\Leftrightarrow x=2008\).
PA
4
20 tháng 7 2018
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\) ( 2/6 = 1/3;2/12=1/6;1/10=2/20;...)
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(1-2.\frac{1}{x+1}=\frac{2007}{2009}\)
\(\frac{2}{x+1}=1-\frac{2007}{2009}\)
\(\frac{2}{x+1}=\frac{2}{2009}\)
=> x +1 = 2009
x= 2008
BO
20 tháng 7 2018
<p>2x[1/6+1/12+1/20+....+1/Xx[x-1]=2007/2008</p><p>2x[1/2x3+1/3x4+1/4x5+....+1/Xx[x-1]=2007/2008</p><p>2x[1/2-1/3+1/3-1/4+1/4-1/5+....+1/[x-1]xX=2007/2008</p><p>1/2-1/x=2007/2008x1/2</p><p>1/2-1/x=2007/4016</p><p>2x[1/2-1/x]=2007/2008</p><p>1/x=1/2-2007/4016 1/x=1/4016.Vay x=4015
J
3
XO
7 tháng 6 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.(x+1)}=\frac{2007}{2009}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2019}\)
Vì 1 = 1
=> x + 1 = 2019
=> x = 2019 - 1
=> x = 2018
NT
1
29 tháng 4 2015
1.1/3+1/6+1/10+...+2/x.(x+1)=2007/2009
=>2/6+2/12+2/20+...+2/x.(x+1)=2007/2009
=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)=2007/2009:2
=>1/2-1/(x+1)=2007/4018
=>1/(x+1)=1/2-2007/4018
=>1/x+1=1/2009
=>x+1=2009
=>x=2009-2008
=>x=1
vậy x=1
NN
3
G
24 tháng 7 2014
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+....+2.\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}\)
\(2.\frac{1}{2}-2.\frac{1}{x}=\frac{2007}{2009}\)
\(\frac{1}{2}-\frac{1}{x}=\frac{2007}{4018}\)
\(\frac{1}{x}=\frac{1}{2}-\frac{2007}{4018}\)
\(\frac{1}{x}=\frac{1}{2009}\)
x=2009
27 tháng 7 2014
22.3+23.4+...+2x.(x−1)=20072009
2.(12−13)+2.(13−14)+....+2.(1x−1−1x)=20072009
2.12−2.1x=20072009
12−1x=20074018
1x=12−20074018
1x=12009
x=2009