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\(A=\frac{1}{5}x\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{10}+\frac{1}{995.1000}\right)\)
\(A=\frac{1}{5}x\left(\frac{1}{5}-\frac{1}{1000}\right)\)
\(A=\frac{1}{5}x\frac{199}{1000}\)
\(A=\frac{199}{5000}\)
Nếu muốn thì thử lại :
\(=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+..+\frac{1}{995}-\frac{1}{1000}\right)...\)
\(=\frac{1}{5}\left(1-\frac{1}{1000}\right)=\frac{1}{5}\cdot\frac{995}{1000}\)
tự tính nốt nha
\(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{45.50}\)
\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\)
\(=\frac{1}{5}-\frac{1}{50}=\frac{10}{50}-\frac{1}{50}=\frac{9}{50}\)
\(=\frac{10-5}{5.10}+\frac{15-10}{10.15}+...+\frac{50-45}{45.50}\)
\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{45}-\frac{1}{50}\)
\(=\frac{1}{5}-\frac{1}{50}=\frac{9}{50}\)
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{1}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)
1/6 + 1/12 + 1/20 + 1/30 + 1/42 + ... + 1/90 + 1/110 = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + ... + 1/9.10 + 1/10.11 = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/9 - 1/10 + 1/10 - 1/11 = 1/2 - 1/11 = 9/22
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
=\(\frac{1}{2}-\frac{1}{11}\)
=\(\frac{9}{22}\)
Đặt biểu thức trên là A ta có:
A = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{48}\)+ \(\frac{1}{96}\)
A x 3 = \(1\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)
A x 3 = \(1\)+ \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)
A x 3 = 2 - \(\frac{1}{32}\)= \(\frac{63}{32}\)
A = \(\frac{63}{32}\): 3 = \(\frac{63}{96}\)
= 1/3 x 5 + 1/5x 7 + 1/7 x 9 +...+1/99 x 101
=1/ 2x (1/3 - 1/5 +1/5 - 1/7 +1/7 - 1/9 + 1/99 - 1/101)
=1/2 x (1/3 - 1/99)
=1/2 x (1/3 - 1/101)
=1/2 x (98/303)
=1/15 + 1/35 + 1/63 +1/99+...+1/9999
=49/303
\(=\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}+0+...+0\)
\(=\frac{98}{303}\)
\(1\cdot\frac{1}{15}\cdot1\frac{1}{16}\cdot1\frac{1}{17}\cdot....\cdot1\frac{1}{2016}\cdot1\frac{1}{2017}\)
\(=\frac{1}{15}\cdot\frac{17}{16}\cdot\frac{18}{17}\cdot....\cdot\frac{2017}{2016}\cdot\frac{2018}{2017}\)
\(=\frac{1}{15}\cdot\frac{1}{16}\cdot2018\)
Dấu "." là dấu nhân nhé bn! phần còn lại bn làm tiếp nha
a) \(\left(\frac{4}{3}-\frac{4}{6}\right)+\left(\frac{4}{6}-\frac{4}{9}\right)+\left(\frac{4}{9}-\frac{4}{10}\right)+\left(\frac{4}{12}-\frac{4}{15}\right)\)
\(=\frac{4}{15}-\frac{4}{3}=\frac{-16}{15}\)
C) bạn chỉ ần bỏ các số giống nhau thôi nhé
= 1
b)
= 1 - \(\frac{1}{2010}\)nha em
#Chúc em học tốt
\(N=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{2005.2010}\)
\(5N=\frac{5}{1.5}+\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2005.2010}\right)\)
\(5N=1+\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2005}-\frac{1}{2010}\right)\)
\(5N=1+\left(\frac{1}{5}-\frac{1}{2010}\right)\)
\(5N=1+\frac{401}{2010}\)
\(5N=\frac{2411}{2010}\)
\(\Rightarrow N=\frac{2411}{10050}\)