1/2-1/3+1/3-1/4+....+1/9-1/10

Ta có 1/2-1/10=2/5

1/2x3 + 1/3x4 + 1/4x5 + 1/5x6 + 1/ 6x7 + 1/7x8 + 1/8x9 + 1/9x10

= 2/5

bạn theo công thức nay nè

\(\frac{n}{a\left(a+n\right)}=\frac{1}{a}-\frac{1}{a+n}\)là giải đc thôi

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\) \(=\) \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

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

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

1/2x3 + 1/3x4 +1/4x5 +...+ 1/9x10

=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

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

\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\)

\(=\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+...+\frac{10}{9.10}-\frac{9}{9.10}\)

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

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

1/2.3+1/3.4+...+1/9.10

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

=1/2-1/10=2/5

Ta có : \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{9.10}\)

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

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

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

=1-1/10

=10/10-1/10

=9/10

Bài làm:

  \(\frac{1}{1\times2}+\frac{1}{2\times3}\)\(+\frac{1}{3\times4}+\frac{1}{4\times5}\)\(+...\frac{1}{9\times10}\)

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

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

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

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

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

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

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

\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{9x10}\)

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

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

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

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{9x10}\)

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

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

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

Tra loi:

1/2x3+1/3x4+1/4x5+1/5x6+1/6x7+1/7x8+1/8x9

=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

=1/2-(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

=1/2-0-0-0-0-0-0-1/9

=1/2-1/9

=7/18

k cho minh nha!

thông điệp nhỏ 

hai tích nếu như ko muốn k

ai tích mình tích lại

cam on Mai Phuong nha!

\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\)

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

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

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

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

\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{10-9}{9.10}\)

\(=\frac{3}{2.3}-\frac{2}{2.3}+\frac{4}{3.4}-\frac{3}{3.4}+\frac{5}{4.5}-\frac{4}{4.5}+...+\frac{10}{9.10}-\frac{9}{9.10}\)

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

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

\(=\frac{5}{10}-\frac{1}{10}\)

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

= 9/10

k nha