=> \(\dfrac{2}{6}\)+\(\dfrac{2}{12}\)+\(\dfrac{2}{20}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\)=\(\dfrac{2011}{2013}\)

=> 2.(\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{2011}{2013}\)

=> 2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{2011}{2013}\)

=> 2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\)) =\(\dfrac{2011}{2013}\)

=>\(\dfrac{x+1-2}{2.\left(x+1\right)}\)=\(\dfrac{2011}{2013}\)

=> \(\dfrac{x-1}{x+1}\)=\(\dfrac{2011}{2013}\)

=> 2013.(x-1) = 2011.(x+1)

=> 2013x-2013= 2011x+2011

=> 2013x -2011x= 2013+2011

=> 2x= 4024

=> x= 2012

Chúc bạn học tốt!Tick cho mk nhé

a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

\(\Rightarrow\)\(2^x+2^x.2+2^x.2^2+2^x.2^3=480\)

\(\Leftrightarrow\)\(2^x\left(1+2+2^2+2^3\right)=480\)

\(\Leftrightarrow\)\(2^x\left(1+2+4+8\right)=480\)

\(\Leftrightarrow\)\(2^x.15=480\)

\(\Rightarrow\)\(2^x=480:15\)

\(\Leftrightarrow2^x=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

Vậy x = 5.

\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)

\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)

\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)

\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)

\(2A=2+3+4+5+6+...+2012+2013+2014\)

\(2A=\dfrac{\left(2+2014\right).2013}{2}\)

\(A=\dfrac{2016.2013}{4}=504.2013\)

\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)

\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)

\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)

\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)

\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)

\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)

\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)

\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\cdot x=\left(1+\dfrac{2011}{2}\right)+\left(1+\dfrac{2010}{3}\right)+...+\left(\dfrac{1}{2012}+1\right)+1\)

\(\Leftrightarrow x\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)=\dfrac{2013}{2}+\dfrac{2013}{3}+...+\dfrac{2013}{2013}\)

=>x=2013

https://hoc24.vn/hoi-dap/question/598367.html

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}\)

\(\Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}\)

\(\Rightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{4020}{2011}\)

\(\Rightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{4020}{2011}\)

\(\Rightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{4020}{2011}:2\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2010}{2011}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2010}{2011}\)

\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{2010}{2011}\)

\(\Rightarrow\dfrac{1}{x+1}=-\dfrac{2009}{4022}\)

\(\Rightarrow4022=-2009\left(x+1\right)\)

\(\Rightarrow4022=-2009x-2009\)

\(\Rightarrow2009x=-2009-4022\)

\(\Rightarrow2009x=-6031\)

\(\Rightarrow x=-\dfrac{6031}{2009}\)

Ta có : \(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right).....\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)

\(=\dfrac{5}{3}.\dfrac{7}{5}....\dfrac{2013}{2011}.\dfrac{2015}{2013}=\dfrac{2015}{3}\)
 

\(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right)\left(1\dfrac{2}{7}\right)...\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)

\(=\dfrac{5}{3}.\dfrac{7}{5}.\dfrac{9}{7}.....\dfrac{2013}{2011}.\dfrac{2015}{2013}\)

\(=\dfrac{2015}{3}\)

\(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right)\left(1\dfrac{2}{7}\right)...\left(1\dfrac{2}{2013}\right)\)

\(=\dfrac{5}{3}.\dfrac{7}{5}.\dfrac{9}{7}...\dfrac{2015}{2013}=\dfrac{2015}{3}\)