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16 tháng 3 2018

B=13/28

10 tháng 2 2019

B=1/7.(555/222+4444/1221+33333/244442+11/330+13/60

B= 1/7.(5/2+4/11+3/22+11/330+13/60)

B= 1/7.(1650/660+240/660+90/660+22/660+143/660)

B=1/7.143/44

B=143/308

k mk nha, k xong bình luận bên dưới mk k lại cho

28 tháng 1 2020

Xin lỗi, mình chỉ làm được câu 1 thôi

\(A=\frac{1}{7}\left(\frac{555}{222}+\frac{4444}{12221}+\frac{33333}{244442}+\frac{11}{330}+\frac{13}{60}\right)\)

\(A=\frac{1}{7}\left(\frac{5.111}{2.111}+\frac{4.1111}{11.1111}+\frac{3.11111}{22.11111}+\frac{11}{11.30}+\frac{13}{60}\right)\)

\(A=\frac{1}{7}\left(\frac{5}{2}+\frac{4}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}\right)\)

\(A=\frac{1}{7}\left[\left(\frac{5}{2 }+\frac{1}{30}+\frac{13}{60}\right)+\left(\frac{4}{11}+\frac{3}{22}\right)\right]\)

\(A=\frac{1}{7}\left[\left(\frac{150}{60}+\frac{2}{60}+\frac{13}{60}\right)+\left(\frac{8}{22}+\frac{3}{22}\right)\right]\)

\(A=\frac{1}{7}\left(\frac{11}{4}+\frac{1}{2}\right)\)

\(A=\frac{1}{7}.\frac{13}{4}\)

\(A=\frac{13}{21}\)

24 tháng 7 2016

a/ (-3,2).\(\frac{-15}{64}\)+(0,8-2\(\frac{4}{5}\)):1\(\frac{23}{24}\)

=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(\(\frac{4}{5}\)-\(\frac{14}{5}\)):\(\frac{47}{24}\)

=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(-2):\(\frac{47}{24}\)

\(\frac{3}{4}\)+\(\frac{-48}{47}\)

=\(\frac{-51}{188}\)

 

25 tháng 7 2016

b/ 1\(\frac{13}{15}\).3.(0,5)\(^2\).3+(\(\frac{8}{15}\)-1\(\frac{19}{60}\)):1\(\frac{23}{24}\)

\(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{8}{15}\)-\(\frac{79}{60}\)):\(\frac{47}{24}\)

\(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)

\(\frac{28}{5}\).\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)

\(\frac{7}{5}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)

\(\frac{21}{5}\)+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)

\(\frac{21}{5}\)+(\(\frac{-2}{5}\))

\(\frac{19}{5}\)

mk làm hơi dài dòng chút 

CHÚC BẠN HỌC TỐT

14 tháng 4 2019

\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)

\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)

\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)

\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)

\(\Rightarrow T=\frac{1004}{1005}\)

14 tháng 4 2019

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)

\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(A=\frac{1}{2}.\frac{2010}{2011}\)

\(\Rightarrow A=\frac{1005}{2011}\)

14 tháng 4 2019

a) C=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)

\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}\)\(=\frac{1.2.3...999}{2.3.4...1000}=\frac{1.\left(2.3.4....999\right)}{\left(2.3.4....999\right).1000}\)\(=\frac{1}{1000}\)

b) Đặt: A=\(1+2+2^2+2^3+...+2^{2008}\)

\(\Leftrightarrow2A=2+2^2+2^3+....+2^{2008}+2^{2009}\)

\(\Leftrightarrow2A-A=2^{2009}-1\)

\(\Leftrightarrow A=2^{2009}-1\)

\(\Rightarrow S=\frac{2^{2009}-1}{1-2^{2009}}\)\(=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=\frac{1}{-1}=-1\)

vậy: S=(-1)