tìm x biết \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=1\frac{2014}{2016}\)
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21 tháng 6 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2016}{2017}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2016}{2017}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2016}{2017}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{2017}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2017}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1008}{2007}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4034}\)
\(\Leftrightarrow x+1=4034\)
\(\Leftrightarrow x=4033\)
Vậy x = 4033
XO
21 tháng 6 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2016}{2017}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2016}{2017}\right)\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2016}{2017}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2016}{1017}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2016}{2017}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2016}{2017}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1008}{2017}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{1008}{2017}\)
=> \(\frac{1}{x+1}=\frac{1}{4034}\)
Vì 1 = 1
=> x + 1 = 4034
=> x = 4034 - 1
=> x = 4033
Lưu ý : Dấu "." là dấu nhân
7 tháng 8 2018
\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
13 tháng 7 2016
a) \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2014}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2014}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2014}\)
\(1-\frac{1}{x+1}=\frac{2015}{2014}\)
\(\frac{1}{x+1}=1-\frac{2015}{2014}\)
\(\frac{1}{x+1}=-\frac{1}{2014}\)
\(x+1=-2014\)
\(x=-2015\)
b) \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2x\left(x+1\right)}=\frac{2984}{1993}\)
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2984}{1993}\)
\(2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2984}{1993}\)
\(2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2984}{1993}\)
\(2\left(1-\frac{1}{x+1}\right)=\frac{2984}{1993}\)
\(1-\frac{1}{x+1}=\frac{1492}{1993}\)
\(\frac{1}{x+1}=\frac{501}{1993}\)
\(501\left(x+1\right)=1993\)không tồn tại số tự nhiên x
HT
3 tháng 7 2018
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
LT
7 tháng 7 2018
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
M
4 tháng 12 2015
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+\right)\left(x+3\right)}+...+\frac{1}{\left(x+2015\right)\left(x+2016\right)}=\frac{1}{x+2016}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+2015}-\frac{1}{x+2016}=\frac{1}{x+2016}\)
\(\frac{1}{x}-\frac{1}{x+2016}=\frac{1}{x+2016}\)
\(\frac{1}{x}-\frac{1}{x+2016}-\frac{1}{x+2016}=0\)
\(\frac{1}{x}-\frac{2x}{x+2016}=0\)
\(\frac{x+2016}{x\left(x+2016\right)}-\frac{2x}{x\left(x+2016\right)}=0\)
\(\frac{x+2016-2x}{x\left(x+2016\right)}=0\Leftrightarrow2016-x=0\Leftrightarrow x=2016\)
MT
3
DP
7 tháng 6 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}\div2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4032}\)
\(\Rightarrow x+1=4032\Rightarrow x=4031\)
Vậy \(x=4031\)
XO
7 tháng 6 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.x+1}\right)=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2032}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{2032}\)
=> \(\frac{1}{x+1}=\frac{1}{2032}\)
Vì 1 = 1
=> x + 1 = 2032
=> x = 2032 - 1
=> x = 2031
x=2016
ta có x=2016 nha^_^,nhớ k cho mình nhé