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\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)
Đặt \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
\(2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(2S-S=1-\frac{1}{2^{10}}\)
\(S=\frac{1024}{1024}-\frac{1}{1024}=\frac{1023}{1024}\)
Vậy \(S=\frac{1023}{1024}\)
P.S: Bạn để \(S=1-\frac{1}{2^{10}}\)vẫn được.
Thực hiện phép tính
a ) \(\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
= \(\frac{2}{5}+\frac{-1}{6}+\frac{-3}{4}+\frac{2}{3}\)
= \(\left(\frac{2}{5}+\frac{-3}{4}\right)+\left(\frac{-1}{6}+\frac{2}{3}\right)\)
= \(\left(\frac{8}{20}+\frac{-15}{20}\right)+\left(\frac{-1}{6}+\frac{4}{6}\right)\)
= \(\left(\frac{8+\left(-15\right)}{20}\right)+\left(\frac{\left(-1\right)+4}{6}\right)\)
= \(\frac{-7}{20}+\frac{1}{2}\)
= \(\frac{-7}{20}+\frac{10}{20}=\frac{\left(7\right)+10}{20}=\frac{3}{20}\)
tk mk nha
đang âm rất nhiều rồi , giúp nha !!!!!
1) \(\frac{2}{5}\cdot\frac{1}{2}+\frac{1}{3}=\frac{1}{5}+\frac{1}{3}=\frac{8}{15}\)
2)\(\left(\frac{5}{7}-\frac{2}{5}\right):\frac{11}{7}=\frac{5}{7}:\frac{11}{7}-\frac{2}{5}:\frac{11}{7}=\frac{5}{7}\cdot\frac{7}{11}-\frac{2}{5}\cdot\frac{7}{11}=\frac{5}{11}-\frac{14}{55}=\frac{1}{5}\)
3)\(\frac{1-\frac{2}{3}+\frac{1}{4}}{2-\frac{1}{3}+\frac{1}{6}}=\frac{12-8+3}{12}:\frac{12-2+1}{6}=\frac{7}{12}\cdot\frac{\frac{6}{11}7}{22}\)
1
1024
1
2014