tìm x biết 1/2.1 +1/2.3+1/3.4 +......+1/x.(x+1) =2014/2015
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25 tháng 3 2021
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}=\frac{2014}{2015}\)
\((1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1})=\frac{2014}{2015}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2014}{2015}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2015}\)
\(\Rightarrow x+1=2015\)
\(\Leftrightarrow x=2014\)
Vậy x=2014
6 tháng 8 2016
1/1.2 +1/2.3 +...+ 1/x(x+1) = 2015/2016
<=> 1-1/2 + 1/2 - 1/3 + ... + 1/x - 1/x+1 = 2015/2016
<=> 1 - 1/x+1 = 2015/2016
<=> 1/x+1 = 1/2016
<=> x + 1 = 2016
<=> x = 2015
LH
6 tháng 8 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(\Leftrightarrow x+1=2016\Rightarrow x=2015\)
TQ
0
21 tháng 10 2023
\(\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2013}{2015}\)
=>\(\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+...+\dfrac{2}{x}-\dfrac{2}{x+1}=\dfrac{2013}{2015}\)
=>\(1-\dfrac{2}{x+1}=\dfrac{2013}{2015}\)
=>\(\dfrac{2}{x+1}=\dfrac{2}{2015}\)
=>x+1=2015
=>x=2014
TQ
1
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
\(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{9}{20}\)
\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\dfrac{1}{2}+0+0+0+...+0-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\dfrac{1}{x+1}=\dfrac{1}{20}\)
\(x+1=20\)
\(x=20-1\)
\(x=19\)
TQ
1
T
12 tháng 8 2023
Có: \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x-1}-\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{9}{20}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{20}\)
\(\Rightarrow x+1=20\Leftrightarrow x=19\)
Lời giải:
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x(x+1)}=\frac{2014}{2015}$
$\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}=\frac{2014}{2015}$
$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}$
$1-\frac{1}{x+1}=\frac{2014}{2015}$
$\frac{1}{x+1}=1-\frac{2014}{2015}=\frac{1}{2015}$
$\Rightarrow x+1=2015$
$\Rightarrow x=2014$