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2 tháng 5 2018
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{x}-\frac{2}{x+1}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{2}-\frac{2}{x+1}=\frac{2005}{2007}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{x+1}=1-\frac{2005}{2007}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2007}\)
\(\Rightarrow x+1=2007\)
\(\Rightarrow x=2006\)
AW
2 tháng 5 2018
\(\frac{1}{2}\cdot\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{2005}{2007}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2005}{4014}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2005}{4014}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2005}{4014}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2005}{4014}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2007}\)
\(\Rightarrow x+1=2007\)
\(x=2007-1\)
\(x=2006\)
ND
2
S
29 tháng 4 2019
1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2005/2007
=> 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 2005/2007
=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x*(x+1) = 2005/2007
=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1) = 2005/2007
=> 2(1/2 - 1/x + 1) = 2005/2007
=> 1/2 - 1/x + 1 = 2005/4014
=> 1/x+1 = 1/2007
=> x + 1 = 2007
=> x = 2006
T
29 tháng 4 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2005}{2007}\)
\(\rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2005}{2007}\)
\(\rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\)
\(\rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\)
\(\rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)
\(\rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2005}{2007}:2\)
\(\rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2005}{2007}:2\) \(\Rightarrow\frac{1}{x+1}=\frac{1}{2007}\)
\(\Rightarrow x+1=2007\rightarrow x=2006\)
Vậy x = 2006.
NT
0
NN
0
OD
5
9 tháng 6 2016
1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005
2 × ( 1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005
1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x+1) = 2003/2005 : 2
1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 × 1/2
1/2 - 1/x+1 = 2003/4010
1/x+1 = 1/2 - 2003/4010
1/x+1 = 2005/4010 - 2003/4010
1/x+1 = 1/2005
=> x+1 = 2005
=> x = 2004
Vậy x = 2004
ai tích mk tích lại cho
9 tháng 6 2016
1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005
2 × ( 1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005
1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x+1) = 2003/2005 : 2
1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 × 1/2
1/2 - 1/x+1 = 2003/4010
1/x+1 = 1/2 - 2003/4010
1/x+1 = 2005/4010 - 2003/4010
1/x+1 = 1/2005
=> x+1 = 2005
=> x = 2004
Vậy x = 2004
ai tích mk tích lại cho
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
Ta có :\(\frac{1}{3}+\frac{1}{6}+..+\frac{2}{x\left(x+1\right)}=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}\)
= 2 x \(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...\frac{1}{x\left(x+1\right)}\right)=2\times\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)\)
= 2 x (\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)
= 2 x (\(\frac{1}{2}-\frac{1}{x+1}\)
Khi đó chỉ cần giải 2 x\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)