A=.....

=\(7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+....+\frac{1}{69.70}\right)=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+.....+\frac{1}{69}-\frac{1}{70}\right)\)

=\(7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)

MẤY PHẦN SAU CX TÁCH MẪU RA RÙI LÀM NHƯ VẬY

TỰ LÀM NHE

\(B=\frac{1}{3\cdot6}+\frac{1}{6\cdot9}+...+\frac{1}{30\cdot33}\)

\(B=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+...+\frac{3}{30\cdot33}\right)\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(B=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)

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

\(C=\left(1-\frac{1}{1\cdot2}\right)+\left(1-\frac{1}{2\cdot3}\right)+...+\left(1-\frac{1}{9\cdot10}\right)\)

\(C=9-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\right)\)

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

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

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

các bạn đều sai hết mà các bạn lại cho những câu hỏi đó đúng 

          A= 1/3 + 1/4-1/4+1/5-1/5+1/6-1/6+1/7-1/7+1/8-1/8+1/9-1/9+1/10

          A=1/3+1/10

          A=13/30

a,\(A=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{8.9}-\frac{1}{9.10}\)

      \(=\frac{1}{12}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

      \(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

       \(=\frac{1}{12}-\frac{1}{4}+\frac{1}{10}=\frac{5}{60}-\frac{15}{60}+\frac{6}{60}=\frac{-1}{15}\) 

Vậy \(A=\frac{-1}{15}\)

\(\frac{1}{9.10}-\frac{1}{8.9}-\frac{1}{7.8}-\frac{1}{6.7}-\frac{1}{5.6}-\frac{1}{4.5}-\frac{1}{3.4}-\frac{1}{2.3}-\frac{1}{1.2}\)

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

\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)+\frac{1}{90}\)

\(=-\left(1-\frac{1}{10}\right)+\frac{1}{90}\)

\(=-\frac{9}{10}+\frac{1}{90}\)

= ...

bn tự tính nha!
 

\(B=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)

\(B=\frac{1}{3}-\frac{1}{10}\)

\(B=\frac{7}{30}\)

sai đề

đề đó là đúng nha bạn

caau hoi tuong tu cungx cos

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

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

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

Đây là tính chứ chứng minh cái gì ? 

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)

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

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

Lập luận: 1/1.2 = 1/1 - 1/2 ; 1/2.3 = 1/2 - 1/3 ; 1/3.4 = 1/3 - 1/4 ; làm tương tự với các số kia.

Ta có: 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10

= 1 - 1/10

= 9/10

\(B=\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}-\frac{1}{8.9}-\frac{1}{9.10}\)

\(\Rightarrow B=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(\Rightarrow B=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow B=\frac{1}{3}-\frac{1}{4}-\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(\Rightarrow B=\frac{1}{12}-\frac{6}{40}\)

\(\Rightarrow B=\frac{-1}{15}\)

de qua

1/5.6+1/6.7+...+1/11.12

ko ghi lại đề

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{210}-\frac{1}{211}+\frac{1}{211}-\frac{1}{212}\)

\(=1-\frac{1}{212}\)

\(=\frac{211}{212}\)

\(\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{211.212}\)

= \(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{211}-\frac{1}{212}\)

= \(\frac{1}{8}-\frac{1}{212}\)

= \(\frac{51}{424}\)

\(\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{210.211}+\frac{1}{211.212}\)

=\(\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{210}-\frac{1}{211}+\frac{1}{211}-\frac{1}{212}\)

=\(\frac{1}{8}-\frac{1}{212}\)=\(\frac{53}{424}-\frac{2}{424}=\frac{51}{424}\)\(\approx0,12028\approx0,12\)

\(M=\frac{1}{9.10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)

yyyyyyyyyyyyyyyyyyyyyyyyyyyy

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