\(S=\dfrac{5}{1.2}+\dfrac{13}{2.3}+\dfrac{25}{3.4}+\dfrac{41}{4.5}+...+\dfrac{181}{9.10}\)

\(S=\dfrac{\left(1^2+2^2\right)}{1.2}+\dfrac{\left(2^2+3^2\right)}{2.3}+...+\dfrac{\left(9^2+10^2\right)}{9.10}\)

\(S=\dfrac{\left\{\left(1-2\right)^2+2.1.2\right\}}{1.2}+\dfrac{\left\{\left(2-3\right)^2+2.2.3\right\}}{2.3}+...+\dfrac{\left\{\left(9-10\right)^2+2.9.10\right\}}{9.10}\)

\(S=\dfrac{\left\{\left(-1\right)^2\right\}}{1.2+2}+\dfrac{\left\{\left(-1\right)^2\right\}}{2.3+2}+...+\dfrac{\left\{\left(-1\right)^2\right\}}{9.10+2}\)

\(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}+2.9\)

\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}+18\)

\(S=1-\dfrac{1}{10}+18\)

\(S=\dfrac{189}{10}\)

Có sai thì đừng ném đá nha tội mình ~~

Câu b, B=\(\dfrac{5}{1\cdot2}+\dfrac{13}{2\cdot3}+\dfrac{25}{3\cdot4}+...+\dfrac{181}{9\cdot10}\)

\(=\left(\dfrac{1}{1\cdot2}+\dfrac{4}{1\cdot2}\right)+\left(\dfrac{1}{2\cdot3}+\dfrac{12}{2\cdot3}\right)+\left(\dfrac{1}{3\cdot4}+\dfrac{24}{3\cdot4}\right)+...+\left(\dfrac{1}{9\cdot10}+\dfrac{180}{9\cdot10}\right)\)=\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)+\left(\dfrac{4}{1\cdot2}+\dfrac{12}{2\cdot3}+...+\dfrac{180}{9\cdot10}\right)\)

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

=\(\dfrac{1}{1}-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-......-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)+\dfrac{1}{10}+\left(2\cdot9\right)\)

=\(1-\dfrac{1}{10}+18\) \(=\dfrac{9}{10}+18\)

=18.9

a, \(\dfrac{\dfrac{3}{2}-\dfrac{2}{5}+\dfrac{1}{10}}{\dfrac{3}{2}-\dfrac{2}{3}+\dfrac{1}{12}}=\dfrac{\dfrac{15}{10}-\dfrac{4}{10}+\dfrac{1}{10}}{\dfrac{18}{12}-\dfrac{8}{12}+\dfrac{1}{12}}=\dfrac{\dfrac{15-4+1}{10}}{\dfrac{18-8+1}{12}}=\dfrac{\dfrac{12}{10}}{\dfrac{11}{12}}=\dfrac{72}{55}\)

Sửa đề : a, \(S=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+\dfrac{3}{4.5}+...+\dfrac{3}{2015.2016}\)

\(=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)

\(=3\left(\dfrac{2016-1}{2016}\right)=3.\dfrac{2015}{2016}=\dfrac{6045}{2016}\)

Câu a) sửa đề: 3/5015.2016 ➜ 3/2015.2016

Giải:

a) S=3/1.2 + 3/2.3 + 3/3.4 +3/4.5 +...+ 3/2015.2016

    S=3.(1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +...+ 1/2015.2016)

    S=3.(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/2015-1/2016)

    S=3.(1-1/2016)

    S=3. 2015/2016

    S=2015/672

b) Mk chưa biết làm nên bạn tự suy nghĩ nhé, xin lỗi!

Ta có:

\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)

`1/(2*3)+1/(3*4)+1/(4*5)+...+1/(9*10)`

`=1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10`

`=1/2-1/10`

`=5/10-1/10`

`=4/10=2/5`

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)

=\(1-\dfrac{1}{5}\)

=\(\dfrac{4}{5}\)

1/1 - 1/2+ 1/2 - 1/3 + 1/3 - 1/4+ 1/4 - 1/5

 = 1/1 - 1/5

= 4/5

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=1-\dfrac{1}{5}=\dfrac{4}{5}\)

\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+\frac{41}{4.5}+...+\frac{181}{9.10}\) \(=\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+\frac{40+1}{20}+...+\frac{180+1}{90}\) 

                                                                        \(=2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+2+\frac{1}{4.5}+...+2+\frac{1}{9.10}\) 

\(=18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=19-\frac{1}{10}\)

\(=\frac{189}{10}\)

chuyện gì ?

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

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

\(=\dfrac{1}{2}-\dfrac{1}{10}\)

\(=\dfrac{2}{5}\)

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

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{10}\)

\(\Rightarrow\dfrac{2}{5}\)