K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

20 tháng 7 2016

\(\frac{2.2}{1.3}x\frac{3.3}{2.4}x\frac{4.4}{3.5}x\frac{5.5}{4.6}x\frac{6.6}{5.7}\)=\(2.\frac{2}{3}.\frac{3}{2}.\frac{3}{4}.\frac{4}{3}.\frac{4}{5}.\frac{5}{4}.\frac{5}{6}.\frac{6}{5}.\frac{6}{7}\)

                                                      \(=2.\frac{6}{7}=\frac{12}{7}\)

20 tháng 7 2016

22/1.3 × 32/2.4 × 42/3.5 × 52/4.6 × 62/5.7

= 2.3.4.5.6/1.2.3.4.5 × 2.3.4.5.6/3.4.5.6.7

= 6 × 2/7

= 12/7

30 tháng 4 2018

\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)

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

\(A=2.\left(1-\frac{1}{7}\right)\)

\(A=2.\frac{6}{7}\)

\(A=\frac{12}{7}\)

30 tháng 4 2018

\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)

\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)

\(A=2.\left(1-\frac{1}{7}\right)\)

\(A=2.\left(\frac{7}{7}-\frac{1}{7}\right)\)

\(A=2.\frac{6}{7}\)

\(A=\frac{12}{7}\)

Chúc bạn học tốt !!! 

21 tháng 7 2017

Bài 1 : 

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{9}{19}\)

\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{9}{19}\)

\(\Leftrightarrow1-\frac{1}{2x+3}=\frac{9}{19}\)

\(\Leftrightarrow\frac{1}{2x+3}=1-\frac{9}{19}\)

\(\Leftrightarrow\frac{1}{2x+3}=\frac{10}{19}\)

\(\Leftrightarrow10.\left(2x+3\right)=19\Leftrightarrow2x+3=\frac{19}{10}\)

\(\Leftrightarrow2x=\frac{19}{10}-3\Leftrightarrow2x=-\frac{11}{10}\)

\(\Leftrightarrow x=-\frac{11}{20}=-0,55\)

Bài 2 : 

\(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{2016}-\frac{1}{2018}\)

\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)

\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(A=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)

30 tháng 7 2020

\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\\ \)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}\)

\(=\frac{4}{8}-\frac{1}{8}\\ =\frac{3}{8}\)

Chúc bn học thiệt giỏi nhé!

19 tháng 7 2017

Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)

\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)

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

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)

1 tháng 9 2015

\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)

=>\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)

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

=>\(S=\frac{1}{2}.\left(1-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{10}\right)\)

=>\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)

=>\(S=\frac{4}{9}-\frac{1}{5}\)

=>\(S=\frac{11}{45}\)

1 tháng 9 2015

lê chí cường dung 

26 tháng 7 2020

Bài làm:

Ta có: \(S=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)

\(>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{9.10}\)

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

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)\(\Rightarrow\frac{2}{5}< S\)

Cái còn lại tự CM