=\(\frac{619}{120}\)

k cho mk nha

\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\)

\(A=1+1+...+1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\)

\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)=9-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

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

\(\frac{1}{2}+\frac{5}{6}+...+\frac{55}{56}\)

\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{56}\right)\)

\(=7.1-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{56}\right)\)

\(=7-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\right)\)

\(=7-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\right)\)

\(=7-\left(1-\frac{1}{8}\right)\)

\(=7-\frac{7}{8}=\frac{49}{8}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}=\frac{1889}{312}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\) \(\frac{89}{90}\)

\(=(1-\frac{1}{2})+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\) \(+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)

\(=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\) 

\(=9-\frac{11}{10}\)

\(=\frac{79}{10}\)

~Học tốt nha~

Đặt : \(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)

\(\Leftrightarrow A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+......+\left(1-\frac{1}{90}\right)\)

\(\Leftrightarrow A=\left(1+1+....+1\right)-\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{90}\right)\)

\(\Leftrightarrow A=9-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)

\(\Leftrightarrow A=9-\left(1-\frac{1}{10}\right)\)

\(\Leftrightarrow A=9-\frac{9}{10}=\frac{81}{90}\)

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)

\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(=9-\left(1-\frac{1}{10}\right)\)

\(=9-\frac{9}{10}=\frac{81}{10}\)

\(\frac{5}{14}\)

A=\(\frac{1}{2}\)+\(\frac{5}{6}\)+\(\frac{11}{12}\)+\(\frac{19}{20}\)+\(\frac{29}{30}\)+\(\frac{41}{42}\)=(1-\(\frac{1}{2}\))+(1-\(\frac{1}{6}\))+(1-\(\frac{1}{12}\))+(1-\(\frac{1}{20}\))+(1-\(\frac{1}{30}\))+(1-\(\frac{1}{42}\))

  =1-1+\(\frac{1}{2}\)+1-\(\frac{1}{2}\)+\(\frac{1}{3}\)+1-\(\frac{1}{3}\)+\(\frac{1}{4}\)+1-\(\frac{1}{4}\)+\(\frac{1}{5}\)+1-\(\frac{1}{5}\)+\(\frac{1}{6}\)+1-\(\frac{1}{6}\)+\(\frac{1}{7}\)

  =(1-1+1+1+1+1+1)+(\(\frac{1}{2}\)-\(\frac{1}{2}\))+(\(\frac{1}{3}\)-\(\frac{1}{3}\))+(\(\frac{1}{4}\)-\(\frac{1}{4}\))+(\(\frac{1}{5}\)-\(\frac{1}{5}\))+(\(\frac{1}{6}\)-\(\frac{1}{6}\))+\(\frac{1}{7}\)

  =5+\(\frac{1}{7}\)=\(\frac{36}{7}\)

đề bài đâu mà tính

đề đâu

mày bị ngu dân à! đề đâu

bn vào câu hỏi tương tự sẽ có chi tiết . Nếu k thì bn hãy để ý mỗi tử đều bé hơn mẫu 1 đơn vị sau đó bn tách ra bằng cách lấy 1 trừ . VD: 5/6 bằng 1  -  1/6 . Đến đó đếm đc 9 chữ số 1 ta lấy 9 làm sbt trừ đi tổng của các ps ta tách đc . Khi đó thì bài toán quá đơn giản rồi . Chúc bn học tốt

(1-1/2)+(1-1/6)+...+(1-1/90)

9+(1/2+1/6+...+1/90)

9+(1/1.2+1/2.3+...+1/9.10)

9+1-9/10=9/1/10=91/10

cho mk nhé

\(A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)

=>\(A=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

=>\(A=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

=>\(A=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

=>\(a=9-\left(1-\frac{1}{10}\right)=\frac{90}{10}-\frac{9}{10}=\frac{81}{10}\)

9 - A = \(1-\frac{1}{2}+1-\frac{5}{6}+1-\frac{11}{12}+..+1-\frac{89}{90}=\frac{1}{2}+\frac{1}{6}+..+\frac{1}{90}\)

         = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)

=> A = \(9-\frac{9}{10}=\frac{81}{10}\)