A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)

tick đi mình giải cho

ngu như con lợn á

\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+.....+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+........+\dfrac{x+2014}{1}\)\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+...+\left(\dfrac{x+1007}{1008}+1\right)=\left(\dfrac{x+1008}{1007}+1\right)+\left(\dfrac{x+1009}{1006}+1\right)+...+\left(\dfrac{x+2014}{1}+1\right)\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+1007}{1008}=\dfrac{x+2015}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+2015}{1008}-\dfrac{x+1008}{1007}-\dfrac{x+2015}{1006}-...-\dfrac{x+2015}{1}=0\)\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1\right)=0\)\(\Leftrightarrow x+2015=0\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1>0\right)\)\(\Leftrightarrow x=-2015\)

Vậy x=-2015

Sửa đề: \(\frac{\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}}{\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}}\)

Ta có: \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2013}-\frac{1}{2014}\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2013}+\frac{1}{2014}-2\left(\frac12+\frac14+\cdots+\frac{1}{2014}\right)\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2013}+\frac{1}{2014}-1-\frac12-\cdots-\frac{1}{1007}\)

\(=\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2014}\)

Ta có: \(\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}\)

\(=\frac{2}{1008\cdot2014}+\frac{2}{1009\cdot2013}+\cdots+\frac{2}{1510\cdot1512}+\frac{1}{1511\cdot1511}\)

\(=2\left(\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{1510\cdot1512}\right)+\frac{1}{1511\cdot1511}\)

\(=\frac{2}{3022}\left(\frac{3022}{1008\cdot2014}+\frac{3022}{1009\cdot2013}+\cdots+\frac{3022}{1510\cdot1512}\right)+\frac{1}{1511\cdot1511}\)

\(=\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{2014}+\frac{1}{1009}+\frac{1}{2013}+\cdots+\frac{1}{1510}+\frac{1}{1512}\right)+\frac{1}{1511}\cdot\frac{1}{1511}\)

\(=\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2013}+\frac{1}{2014}\right)\)

Ta có: \(\frac{\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}}{\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}}\)

\(=\frac{\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2014}}{\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2013}+\frac{1}{2014}\right)}\)

\(=1:\frac{1}{1511}=1511\)