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\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{1017.2019}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2017\cdot2019}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\)
a) \(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}+\frac{3}{418}+\frac{3}{550}\)
= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}+\frac{3}{19.22}+\frac{3}{22.25}\)
= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)
= \(\frac{1}{1}-\frac{1}{25}\)
= \(\frac{24}{25}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n+1\right).\left(2n+3\right)}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)
= \(\frac{1}{1}-\frac{1}{2n+3}\)
= \(\frac{2n+2}{2n+3}\)
c) \(\frac{7+\frac{7}{13}-\frac{7}{48}+\frac{7}{95}}{15+\frac{15}{13}-\frac{15}{48}+\frac{15}{95}}-\frac{7070707}{15151515}\)
= \(\frac{7\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}{15\left(1+\frac{1}{13}-\frac{1}{48}+\frac{1}{95}\right)}-\frac{7.1010101}{15.1010101}\)
= \(\frac{7}{15}-\frac{7}{15}\)
= 0
a) =1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101
=100/101
b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5
=(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101).2,5
=(1-1/101).2,5
=100/101.2,5
=250/101
dấu / là phần nhé. bạn có thể xem bài có dấu phần ở : Câu hỏi của Nguyễn Thị Hoài Anh
A)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
=1-\(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=1-\(\frac{1}{101}\)
=\(\frac{100}{101}\)
B) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{1}{99.101}\)
=5.(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{2}{2}.\)(\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{1}{2}\).(\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\))
=5.\(\frac{1}{2}\).(1-\(\frac{1}{3}\)+\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
=5.\(\frac{1}{2}\).(1-\(\frac{1}{101}\))
=\(\frac{5}{2}.\frac{100}{101}=\frac{250}{100}\)
Chúc bạn học tốt
\(\frac{8}{15}.\frac{3}{4}+\frac{8}{15}.0\)
\(=\frac{2}{5}+0\\ =\frac{2}{5}\)
\(a,=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(b,=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)
a,\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)=\frac{2}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)=1.\frac{99}{100}=\frac{99}{100}\)
\(M=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(\Rightarrow M=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\Rightarrow2M=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(\Rightarrow2M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow2M=\frac{1}{3}-\frac{1}{51}\)
\(\Rightarrow2M=\frac{16}{51}\)
\(\Rightarrow M=\frac{8}{51}\)
\(N=\frac{-5}{1.3}+\frac{-5}{3.5}+...+\frac{-5}{2013.2015}\)
\(\Rightarrow N=-\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)\)
\(\Rightarrow N=-\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(\Rightarrow N=-\frac{5}{2}\left(1-\frac{1}{2015}\right)\)
\(\Rightarrow N=-\frac{5}{2}.\frac{2014}{2015}\)
\(\Rightarrow N=-\frac{1007}{403}\)
\(S=\frac{1}{1.2}+\frac{1}{3.4}+.........+\frac{1}{199.200}\)
a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
a ) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
\(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
\(=\left(\frac{4}{20}+\frac{3}{20}\right)+\left(\frac{8}{21}+\frac{2}{21}-\frac{10}{21}\right)+\left(\frac{2}{5}-\frac{3}{5}\right)\)
\(=\frac{7}{20}+0+\frac{-1}{5}=\frac{7-4}{20}=\frac{3}{20}\)
b ) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)
\(=\frac{21}{23}+\frac{-21}{23}+\frac{-125}{143}\)
\(=0+\frac{-125}{143}=-\frac{125}{143}\)
bài 2
a \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)
=\(1-\frac{1}{2004}=\frac{2003}{2004}\)
Bài làm
\(S_1=\frac{15}{1.3}+\frac{15}{3.5}+\frac{15}{5.7}+...+\frac{15}{2017.2019}\)
\(S_1=15.\frac{1}{1.3}+15.\frac{1}{3.5}+15.\frac{1}{5.7}+...+15.\frac{1}{2017.2019}\)
\(S_1=15.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\right)\)
\(S_1=15.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(S_1=15.\left(1-\frac{1}{2019}\right)\)
\(S_1=15.\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(S_1=15.\frac{2018}{2019}\)
\(S_1=\frac{2018}{673}\)
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