không quy đồng hãy tính hợp lí các tổng sau :
a) A = \(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
b) B = \(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
K
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19 tháng 7 2018
\(A=-\frac{1}{20}+-\frac{1}{30}+-\frac{1}{42}+...+-\frac{1}{90}\)
\(\Leftrightarrow A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(A=-\frac{3}{20}\)
5 tháng 3 2016
a) A = \(\frac{1}{5}\) - \(\frac{1}{4}\)+ \(\frac{1}{6}\)- \(\frac{1}{5}\)+ \(\frac{1}{7}\)-\(\frac{1}{6}\)+\(\frac{1}{8}\)-\(\frac{1}{7}\)+\(\frac{1}{9}\)- \(\frac{1}{8}\)+ \(\frac{1}{10}\)- \(\frac{1}{9}\)
= \(\frac{-1}{4}\)+\(\frac{1}{10}\)= \(\frac{-6}{40}\)= \(\frac{-3}{20}\)
b) B = \(\frac{5}{2.1}\)+ \(\frac{1}{11}\)(4 + \(\frac{3}{2}\)) + \(\frac{1}{2.15}\)(1 + \(\frac{13}{2}\))
= \(\frac{5}{2.1}\)+ \(\frac{1}{11}\).\(\frac{11}{2}\)+ \(\frac{1}{2.15}\).\(\frac{15}{2}\)
= \(\frac{5}{2}\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)= 3 + \(\frac{1}{4}\)= \(\frac{13}{4}\)
19 tháng 11 2021
\(a,\Rightarrow A=-1\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{9.10}\right)\)
\(\Rightarrow A=-1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(\Rightarrow A=-1\left(\dfrac{1}{4}-\dfrac{1}{10}\right)\)
\(\Rightarrow A=\dfrac{-3}{20}\)
19 tháng 11 2021
Bài 2:
\(a,\dfrac{1717}{8585}=\dfrac{1717:1717}{8585:1717}=\dfrac{1}{5};\dfrac{1313}{5151}=\dfrac{1313:101}{5151:101}=\dfrac{13}{51}\\ \dfrac{1}{5}=\dfrac{51}{255}< \dfrac{65}{255}=\dfrac{13}{51}\\ \Rightarrow\dfrac{1717}{8585}< \dfrac{1313}{5151}\)
\(b,\dfrac{201201}{202202}=\dfrac{201201:1001}{202202:1001}=\dfrac{201}{202}=\dfrac{201\cdot1001001}{202\cdot1001001}=\dfrac{201201201}{202202202}\)
NN
3
21 tháng 4 2017
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{15}=\frac{13}{30}\)
A=\(\frac{13}{30}.7=\frac{91}{30}\)
25 tháng 3 2016
A= \(\frac{-1}{4\cdot5}+\frac{-1}{5\cdot6}+\frac{-1}{6\cdot7}+\frac{-1}{7\cdot8}+\frac{-1}{8\cdot9}+\frac{-1}{9\cdot10}\)
=\(-1\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)
=\(-1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=\(-1\left(\frac{1}{4}-\frac{1}{10}\right)\)
=\(-1\cdot\frac{3}{20}\)
=\(\frac{-3}{20}\)
=\(\frac{-1}{20}\)
13 tháng 7 2018
\(A=-\frac{1}{20}+-\frac{1}{30}+-\frac{1}{42}+-\frac{1}{56}+-\frac{1}{72}+-\frac{1}{90}\)
\(\Rightarrow A=-1\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}\right)\)
\(A=-1\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(\Rightarrow A=-\frac{3}{20}\)
D
13 tháng 7 2018
\(A=\frac{-1}{20}-\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(A=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(A=\frac{-3}{20}\)
#
\(a,A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)
\(=\frac{-1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}-...-\frac{1}{9}+\frac{1}{10}\)
\(=-\frac{1}{4}+\frac{1}{10}\)
\(=-\frac{3}{20}\)
\(b,B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{11.7}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-....-\frac{1}{28}\)
\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
a) \(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
\(\Rightarrow-1.A=\frac{1}{20}+\frac{1}{30}+........+\frac{1}{90}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+........+\frac{1}{9.10}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{9}-\frac{1}{10}=\frac{1}{4}-\frac{1}{10}=\frac{3}{20}\)
\(\Rightarrow A=\frac{3}{20}:\left(-1\right)=\frac{-3}{20}\)
b) \(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(\Rightarrow\frac{1}{7}B=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(\Rightarrow B=\frac{13}{28}:\frac{1}{7}=\frac{13}{28}.7=\frac{13}{4}\)