A= \(\frac{-1}{4\cdot5}+\frac{-1}{5\cdot6}+\frac{-1}{6\cdot7}+\frac{-1}{7\cdot8}+\frac{-1}{8\cdot9}+\frac{-1}{9\cdot10}\)

=\(-1\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)

=\(-1\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)

=\(-1\left(\frac{1}{4}-\frac{1}{10}\right)\)

=\(-1\cdot\frac{3}{20}\)

=\(\frac{-3}{20}\)

=\(\frac{-1}{20}\)

phân tích mẫu: 20=4.5 , 30= 5.6 , 42=6.7 tương tự rồi tách cả phân số là được

\(A=-\frac{1}{20}+-\frac{1}{30}+-\frac{1}{42}+-\frac{1}{56}+-\frac{1}{72}+-\frac{1}{90}\)

\(\Rightarrow A=-1\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{9.10}\right)\)

\(A=-1\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(\Rightarrow A=-\frac{3}{20}\)

\(A=\frac{-1}{20}-\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

\(A=-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(A=-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=-\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(A=\frac{-3}{20}\)

#

A=\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)\(\frac{-1}{90}\)

A=\(-1-\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\)

A=\(-1-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

A=\(-1-\left(\frac{1}{4}-\frac{1}{10}\right)\)

A=\(-1-\frac{3}{20}\)

A=\(-1\frac{3}{20}\)

b,Ta có 

\(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)

\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)

\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)

\(\Rightarrow P>Q\)

\(A=\frac{-10}{20}+\frac{-10}{30}+\frac{-10}{42}+\frac{-10}{56}+\frac{-10}{72}+\frac{-10}{90}+\frac{-10}{110}\)

\(=-10\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)\)

\(=-10\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)

\(=-10\left(\frac{1}{4}-\frac{1}{11}\right)\)

\(=\frac{-35}{22}\)

Không liên quan nhưng tui cũng là fan Jungkook

bằng -223/20

A = \(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

\(A=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)

\(A=\frac{-1}{4}-\frac{-1}{5}+\frac{-1}{5}-\frac{-1}{6}+\frac{-1}{7}-\frac{-1}{8}+\frac{-1}{8}-\frac{-1}{9}+\frac{-1}{9}-\frac{-1}{10}\)

\(A=\frac{-1}{4}-\frac{-1}{10}\)

\(A=\frac{-3}{20}\)

\(a,A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)

\(=\frac{-1}{4}+\frac{1}{5}-\frac{1}{5}+\frac{1}{6}-...-\frac{1}{9}+\frac{1}{10}\)

\(=-\frac{1}{4}+\frac{1}{10}\)

\(=-\frac{3}{20}\)

\(b,B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

\(\frac{B}{7}=\frac{5}{2.7}+\frac{4}{11.7}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-....-\frac{1}{28}\)

\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

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) \(B=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)

\(\Rightarrow\frac{1}{7}B=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)

\(=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)

\(\Rightarrow B=\frac{13}{28}:\frac{1}{7}=\frac{13}{28}.7=\frac{13}{4}\)

                                                                  Bài giải

a, \(A=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

\(-1A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

\(-1A=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(-1A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(-1A=\frac{1}{4}-\frac{1}{10}\)

\(-1A=\frac{3}{20}\)

\(A=\frac{3}{20\text{ }}\text{ : }\left(-1\right)=-\frac{3}{20}\)

b, \(B=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}+\frac{13}{15\cdot4}\)

\(\frac{1}{7}B=\frac{5}{2\cdot7}+\frac{4}{7\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)

\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)

\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{28}\)

\(\frac{1}{7}B=\frac{13}{28}\)

\(B=\frac{13}{28}\text{ : }\frac{1}{7}=\frac{13}{4}\)

\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90} \)\(\frac{-1}{90}\)

\(=\frac{-1}{4.5}+\frac{-1}{5.6}+\frac{-1}{6.7}+\frac{-1}{7.8}+\frac{-1}{8.9}+\frac{-1}{9.10}\)

\(=\frac{-1}{4}-\frac{-1}{5}+\frac{-1}{5}-\frac{-1}{6}+\frac{-1}{6}-\frac{-1}{7}+\frac{-1}{7}-\frac{-1}{8}+\frac{-1}{8}-\frac{-1}{9}+\frac{-1}{9}-\frac{-1}{10}\)

\(=\frac{-1}{4}-\frac{-1}{10} \)\(=\frac{-3}{20}\)

\(\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)

= \(\frac{-1}{4}-\frac{-1}{10}\)

=\(\frac{-3}{20}\)

\(A=-\frac{1}{20}+-\frac{1}{30}+-\frac{1}{42}+...+-\frac{1}{90}\)

\(\Leftrightarrow A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(A=\left(-1\right)\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(A=-\frac{3}{20}\)