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\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
Ai giúp mình câu này với
(1+1/2+1/3+...+1/2012+1/2013) .x +2013 = 2014+2015/2+...+4025/2012+4026/2013
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B
\(\Rightarrow\) \(\dfrac{A}{B}\)=2015
\(\left(\frac{2013}{2011}+\frac{2014}{2012}+\frac{2015}{2013}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)=0\)