DH
2
K
Khách
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CT
5
TT
0
8 tháng 8 2017
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+..........+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2014}{2016}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+.............+\dfrac{2}{x\left(x+1\right)}=\dfrac{2014}{2016}\)
\(\Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2014}{2016}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2014}{2016}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2014}{2016}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1007}{2016}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2016}\)
\(\Leftrightarrow x+1=2016\)
\(\Leftrightarrow x=2015\left(tm\right)\)
Vậy ...........
KT
3
HG
2
23 tháng 7 2015
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2014}{2015}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1007}{2015}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1007}{2015}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4030}\)
=>x+1=4030
=>x=4029
vậy x=4029
HG
0
AH
Akai Haruma
Giáo viên
30 tháng 9
Lời giải:
$1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x(x+1)}=\frac{2014}{2015}$
$\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x(x+1)}=\frac{2014}{2015}$
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}=\frac{1007}{2015}$
$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1007}{2015}$
$1-\frac{1}{x+1}=\frac{1007}{2015}$
$\frac{1}{x+1}=1-\frac{1007}{2015}=\frac{1008}{2015}$
$\Rightarrow x+1=\frac{2015}{1008}$
$\Rightarrow x=\frac{1007}{1008}$
\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x(x+1)}=\frac{2014}{2015}$
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2014}{2016}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2014}{2016}\)
\(\Leftrightarrow\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2014}{2016}\)
\(\Leftrightarrow\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1007}{2016}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1007}{2016}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1007}{2016}\)\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2016}\)
\(\Leftrightarrow x+1=2016\Leftrightarrow x=2015\)
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