Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(2\right)K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(K=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(K=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\left(3\right)L=\frac{5}{1\cdot4}+\frac{5}{4\cdot7}+\frac{5}{7\cdot10}+...+\frac{5}{100\cdot103}\)
\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\cdot\left(\frac{1}{1}-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{510}{309}=\frac{170}{103}\)
Trả lời:
2,\(K=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(K=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(K=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(K=\frac{1}{2}-\frac{1}{100}\)
\(K=\frac{49}{100}\)
3,\(L=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(L=\frac{5}{3}\times\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)
\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\times\left(\frac{1}{1}-\frac{1}{103}\right)\)
\(L=\frac{5}{3}\times\frac{102}{103}\)
\(L=\frac{170}{103}\)
Học tốt
\(\frac{-3}{5}+\frac{28.43}{5.56}+\frac{28.5}{5.24}-\frac{28.21}{5.63}\)
=\(\frac{-3}{5}+\frac{2.2.7.43}{5.2.2.7.2}+\frac{2.2.7.5}{2.2.2.3.5}-\frac{2.2.7.3.7}{5.7.3.3}\)
=\(\frac{-3}{5}+\frac{43}{5.2}+\frac{7}{2.3}-\frac{28}{15}\)
=\(\frac{-3}{5}+\frac{43}{10}+\frac{7}{6}-\frac{28}{15}\)
=\(\frac{-18}{30}+\frac{129}{30}+\frac{35}{30}-\frac{56}{30}\)
=\(\frac{-18+129+35-56}{30}\)
=\(\frac{90}{30}=3\)
Chúc bn học tốt
\(M=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(\Rightarrow M=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\Rightarrow2M=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(\Rightarrow2M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(\Rightarrow2M=\frac{1}{3}-\frac{1}{51}\)
\(\Rightarrow2M=\frac{16}{51}\)
\(\Rightarrow M=\frac{8}{51}\)
\(N=\frac{-5}{1.3}+\frac{-5}{3.5}+...+\frac{-5}{2013.2015}\)
\(\Rightarrow N=-\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)\)
\(\Rightarrow N=-\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(\Rightarrow N=-\frac{5}{2}\left(1-\frac{1}{2015}\right)\)
\(\Rightarrow N=-\frac{5}{2}.\frac{2014}{2015}\)
\(\Rightarrow N=-\frac{1007}{403}\)
A= -8/5: (1+2/3)
= -8/5:5/3
= -8/5.3/5
= -24/25
B= 7/5.15/49-22/15: 11/5
= 3/7-2/5
= 1/35
\(A=-1,6:\left(1+\frac{2}{3}\right)\)
\(A=-1,6:\frac{5}{3}\)
\(A=-\frac{24}{25}\)
\(B=1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(B=\frac{135}{98}-\frac{2}{3}\)
\(B=\frac{209}{294}\)
\(A=-1,6:\left(1+\frac{2}{3}\right)\)
\(A=\frac{-8}{5}:\frac{5}{3}=-\frac{8}{5}.\frac{3}{5}=\frac{-24}{25}\)
\(B=\frac{7}{5}.\frac{15}{29}-\left(\frac{4}{5}+\frac{2}{3}\right):\frac{11}{5}\)
\(B=\frac{21}{29}-\left(\frac{12}{15}+\frac{10}{15}\right).\frac{5}{11}=\frac{21}{29}-\frac{22}{15}.\frac{5}{11}=\frac{21}{29}-\frac{2}{3}\)
\(B=\frac{63}{87}-\frac{58}{87}=\frac{5}{87}\)
=> ( \(\frac{-5}{2.3}+\frac{-5}{3.4}+\frac{-5}{4.5}+.....+\frac{-5}{9.10}\)):(\(\frac{1}{4}+\frac{7}{6}-\frac{7}{3}\)):\(\frac{24}{33}\)
=>\(\frac{-5}{1}.\)(\(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)):(\(\frac{3}{12}+\frac{14}{12}-\frac{28}{12}\)):\(\frac{24}{33}\)
=>-5.(\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)):(\(\frac{-11}{12}\)):\(\frac{24}{33}\)
=>-5.(\(\frac{1}{2}-\frac{1}{10}\)):\(\frac{-11}{10}\):\(\frac{24}{33}\)
=>-5.\(\frac{2}{5}\):\(\frac{-11}{10}:\frac{24}{33}\)
=>2:\(\frac{-11}{10}:\frac{24}{33}\)
=>\(\frac{-5}{2}\)
\(\frac{14}{45}\)thi phai