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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 \(x=4031\)

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

9 tháng 8 2016

bó tay

26 tháng 4 2015

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015.\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\)

=> x = 2015

 

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x+2015=\frac{2016}{1}+\frac{2017}{2}+\frac{2018}{3}+...+\frac{4030}{2015}\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)

\(\Rightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right).x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}=2015.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)\(\Rightarrow x=2015\)

5 tháng 5 2015

giải rồi mà cũng hỏi                

30 tháng 4 2015

 

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x+2015=\frac{2016}{1}+\frac{2017}{2}+...+\frac{4030}{2015}\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\left(\frac{2016}{1}-1\right)+\left(\frac{2017}{2}-1\right)+...+\left(\frac{4030}{2015}-1\right)\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=\frac{2015}{1}+\frac{2015}{2}+...+\frac{2015}{2015}\)

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)x=2015.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)

\(\Rightarrow x=2015\)

Bạn có thể tham khảo nhé!^-^

30 tháng 4 2015

Nhi giải nhanh đi                         

17 tháng 4 2018

\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)

\(\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}\right)=\frac{1}{2}.\frac{2015}{2017}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2015}{4034}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{4034}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4034}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)

\(\frac{1}{x+1}=\frac{1}{2017}\)

\(\Rightarrow\)x+1=2017

\(\Rightarrow\)x=2017-1

        x=2016

Vậy x=2016

Chúc bạn học tốt+-*/

17 tháng 4 2018

x = -2018 nha bạn 

6 tháng 4 2018

\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)

=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)

=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)

=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)=  \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)

=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

=> \(x=10\)

b) Tương tự câu a