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\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
\(A=2014.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\right)\)
\(A=2014.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1007.2013}\right)\)
\(A=2.2014.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2014}\right)\)
\(A=2.2014.\frac{2013}{2014}\)
\(A=\frac{2.2014.2013}{2014}\)
\(A=2.2013\)
\(A=4026\)
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
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: \(\frac{\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2013\cdot2014}}{\frac{1}{1008\cdot2014}+\frac{1}{1009\cdot2013}+\cdots+\frac{1}{2014\cdot1008}}\)
\(=\frac{\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2014}}{\frac{1}{1511}\left(\frac{1}{1008}+\frac{1}{1009}+\cdots+\frac{1}{2013}+\frac{1}{2014}\right)}\)
\(=1:\frac{1}{1511}=1511\)
\(1.2.3.4..............2015-1.2.3.4..............2014-1.2.3.4.........2013.2014^2\)
\(=1.2.3........2013.\left(2014.2015-2014-2014^2\right)\)
\(=1.2.3..........2013.\left[2014.\left(2015-1-2014\right)\right]\)
\(=1.2.3............2013.\left(2014.0\right)\)
\(=1.2.3.........2013.0\)
=0
Chúc bn học tốt
2014+(2014/1+2)+(2014/1+2+3)+...+(2014/1+2+3+...+2013)
=2014*(1+(1/1+2)+(1/1+2+3)+...+( 1/1+2+3+...+2013))
=2014*(1+(1/3)+(1/6)+....+(1/2027091)
=2014*2*((1/+(1/2*3)+(1/3*4).....+(1/2013*2014))
=2014*2*(1/1-1/2+1/2-1/3+1/3-1/4+.....+1/2013-1/2014)
=2014*2*(1-1/2014)
=2*(2014*2013/2014)
=2*2013
=4026
Cuối cùng cũng giải được.