Các câu hỏi tương tự
Tim x
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
2/ tim x
\(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7} +\frac{x+2018}{8}\)
3/ tim x
\(\frac{1}{3}+\frac{1}{6}+\frac{99}{101}+\frac{1}{15}+... +\frac{1}{x\left(2x+1\right)}=\frac{1}{10}\)
so sánh 2 số A và B nếu
\(A=-\frac{1}{2018}-\frac{3}{2017^2}-\frac{5}{2017^3}-\frac{7}{2017^4};B=\frac{-1}{2018}-\frac{7}{2017^2}-\frac{5}{2017^3}-\frac{3}{2017^4}\)
So sánh A và B nếu
\(A=\frac{-1}{2018}-\frac{3}{2017^2}-\frac{5}{2017^3}-\frac{7}{2017^4}\)
\(B=\frac{-1}{2018}-\frac{7}{2017^2}-\frac{5}{2017^3}-\frac{3}{2017^4}\)
\(\frac{x-2017}{5}-\frac{x-2017}{6}=\frac{x-2017}{7}-\frac{x-2018}{8}\)
s=1-\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
và p=\(\frac{1}{1007}+\frac{1}{1008}+....+\frac{1}{2012}+\frac{1}{2013}\)
tính(S-P)^2013
đề thi chọn học sinh giỏi môn toán, lớp 7, tỉnh bắc giang năm học 2012-2013
\(\frac{\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+.....+\frac{1}{2013}}}{ }\)
Rút gọn \(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(M=\hept{\begin{cases}0.4-\frac{2}{9}+\frac{2}{11}-\frac{1}{3}-0.25+\frac{1}{5}\\1,4-\frac{7}{9}+\frac{7}{11}-1\frac{1}{6}-0.875+0.7\end{cases}}\left\{\right\}:\frac{2012}{2013}\)
Tính\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+\frac{2010}{4}+...+\frac{1}{2013}}\)