tim x : 1/3+1/6 +1/10 + .... +2/x(x+1) = 2016/2018
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8 tháng 7 2017
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
30 tháng 3 2017
\(\frac{1}{3}\) + \(\frac{1}{6}\) + \(\frac{1}{10}\) + ... + \(\frac{1}{x\left(x+1\right):2}\)
= \(\left(1-\frac{1}{2018}\right)-\frac{1}{2018}\)
= \(\frac{2017}{2018}-\frac{1}{2018}\)
= \(\frac{2016}{2018}=\frac{1008}{1009}\)
27 tháng 12 2017
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2016}{2018}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{504}{1009}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{504}{1009}\)
\(\frac{1}{x+1}=\frac{1}{2018}\)
\(\Rightarrow x+1=2018\)
\(\Rightarrow x=2017\)
TD
2
KT
18 tháng 8 2018
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2016}{2018}\)
<=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1008}{1009}\)
<=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1008}{1009}\)
<=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{504}{1009}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{504}{1009}\)
<=> \(\frac{1}{x+1}=\frac{1}{2018}\)
=> \(x+1=2018\)
<=> \(x=2017\)
1 tháng 1 2018
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.........+\dfrac{1}{x\left(x+1\right)}=\dfrac{2016}{2018}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2017}\)
\(\Leftrightarrow2\left(\dfrac{1}{6}+\dfrac{1}{12}+........+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2016}{2018}\)
\(\Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2016}{2018}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1008}{2018}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2018}\)
\(\Leftrightarrow x+1=2018\)
\(\Leftrightarrow x=2017\)
Vậy ...
5 tháng 7 2018
Chỉ làm bài khó thôi nhé:::::::::::::::
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2016}{2018}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x.\left(x+1\right)}=\frac{2016}{2018}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2016}{2018}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{1013}{2018}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1013}{2018}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1013}{2018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2018}\Rightarrow x+1=2018\Rightarrow x=2017\)
18 tháng 6 2016
a) \(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\Leftrightarrow\orbr{\begin{cases}x\ge\frac{3}{2}\mid:2x-3=\frac{1}{4}\Rightarrow2x=\frac{13}{4}\Rightarrow x=\frac{13}{8}\left(TM\right)\\x< \frac{3}{2}\mid:3-2x=\frac{1}{4}\Rightarrow2x=\frac{11}{4}\Rightarrow x=\frac{11}{8}\left(TM\right)\end{cases}.}\)
b) \(\Leftrightarrow\left|x-1\right|=\frac{3}{4}\Leftrightarrow\orbr{\begin{cases}x\ge1\mid:x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\left(TM\right)\\x< 1\mid:1-x=\frac{3}{4}=>x=\frac{1}{4}\left(TM\right)\end{cases}}\)
c) \(\frac{3}{5\left(x-\frac{5}{6}\right)}-\frac{1}{2\left(\frac{3}{2}-1\right)}=-\frac{1}{4}\Leftrightarrow\frac{3}{\frac{5\left(6x-5\right)}{6}}-\frac{1}{2\cdot\frac{1}{2}}=-\frac{1}{4}\Leftrightarrow\frac{18}{5\left(6x-5\right)}=-\frac{1}{4}+1\)
\(\Leftrightarrow\frac{18}{5\left(6x-5\right)}=\frac{3}{4}\Leftrightarrow6x-5=\frac{24}{5}\Leftrightarrow6x=\frac{49}{5}\Leftrightarrow x=\frac{49}{30}\)
d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\Leftrightarrow2\cdot\frac{x+1-2}{2\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow\frac{x-1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow2016x-2016=2015x+2015\Leftrightarrow x=2015+2016=4031\)
Vậy x = 4031.
MV
11 tháng 5 2017
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\cdot\left(x+1\right):2}=\dfrac{2016}{2018}\\ \dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\cdot\left(x+1\right)}=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2018}\\ 2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}:2\\ \dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1008}{2018}\\ \dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{1008}{2018}\\ \dfrac{1}{x+1}=\dfrac{1}{2018}\\ \Leftrightarrow x+1=2018\\ x=2018-1\\ x=2017\)
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2017}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2017}\)
\(\Leftrightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+..........+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2018}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2016}{2018}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1008}{2018}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2018}\)
\(\Leftrightarrow x+1=2018\)
\(\Leftrightarrow x=2017\)
Vậy ..
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2016}{2018}\)
\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1008}{1009}\)
\(2.\left(\dfrac{1}{2}-\dfrac{1}{x-1}\right)\) = \(\dfrac{1008}{1009}\)
\(\dfrac{1}{2}-\dfrac{1}{x-1}=\dfrac{504}{1009}\)
\(\dfrac{1}{x-1}=\dfrac{1}{2018}\)
\(x-1=2018\)
\(x=2019\)