D2
1
K
Khách
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DN
0
X
9 tháng 2 2021
ĐKXĐ: \(x\ne0;x\ne-1\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2008}{2010}\)
\(\Leftrightarrow2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2008}{2010}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)(Biết công thức này chứ?)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)
\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2008}{2010}\)
\(\Leftrightarrow\dfrac{x-1}{x+1}=\dfrac{2008}{2010}\Leftrightarrow2010x-2010=2008x+2008\Leftrightarrow x=2009\left(tm\right)\)
Vậy x = 2009
15 tháng 1 2017
1+2.( 1/2-1/3+1/3-1/4+....+1/(x-1)-1/x+1)=3/2
1+2.(1/2-1/x+1)=3/2
1-2/x+1=3/2-1
tự tính
NK
0
28 tháng 4 2021
d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)
\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)
Vậy: x=-1
AD
2
HT
9 tháng 8 2015
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{2001}:2=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}=\frac{1}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
9 tháng 8 2015
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}:\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
TL
1
15 tháng 4 2020
đặt A = 1.2. + 2.3 + 3.4 + ... + 49.50
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 49.50.3
3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50
3A = 49.50.51
A = 41650
Thay vào ta được
41650 + 1/2x = 40642
=> 1/2x = 1008
=> x = 2016
Đặt \(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2.}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\Rightarrow\frac{S}{2}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2001}\)
\(\Rightarrow\)x+1=2001
x=2000
Vậy x=2000.