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\(S.2=\frac{2}{2.4}+\frac{2}{4.6}+...........+\frac{2}{98.100}\)
\(S.2=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+.....+\frac{1}{98}-\frac{1}{100}\)
\(S.2=\frac{1}{2}-\frac{1}{100}\)
\(S.2=\frac{49}{100}\)
\(S=\frac{49}{100}:2\)
\(S=\frac{49}{200}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
\(=\frac{1}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+...+\frac{2}{12\times14}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{2}\times\frac{3}{7}=\frac{3}{14}\)
\(A=\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+...+\frac{1}{2012\times2014}\)
\(=\frac{1}{2}\times(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2012\times2014})\)
\(=\frac{1}{2}\times(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2012}-\frac{1}{2014})\)
\(=\frac{1}{2}\times(\frac{1}{2}-\frac{1}{2014})\)
\(=\frac{1}{2}\times(\frac{1007}{2014}-\frac{1}{2014})\)
\(=\frac{1}{2}\times\frac{503}{1007}\)
\(=\frac{503}{2014}\)
Ta có ; \(\frac{1}{2}=\frac{1007}{2014}\)
Vậy A bé hơn B
Chúc bạn học tốt
Ta có: \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
= \(\frac{1}{2}+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+....+\left(\frac{1}{64}-\frac{1}{128}\right)\)
=\(\frac{1}{2}+\frac{1}{2}-\frac{1}{128}\)
\(=1-\frac{1}{128}=\frac{127}{128}\)
Lời giải :
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
ko chép lại đề :
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ......... + \(\frac{1}{98}\)- \(\frac{1}{99}\)+ \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
A) \(\frac{1}{6}\) = 0,1666666665
B) 0,1666669167
\(\frac{1}{6}\) < \(\frac{111111}{666665}\)
Bạn lấy tử chia cho mẫu là ra
lấy 2 S