=1/2-1/3+1/3-1/4+.....+1/49+1/50

=1/2-1/50

=25/50-1/50

=24/50

=12/25

\(\frac{1}{2x3}\)+   \(\frac{1}{3x4}\)+  ...  +  \(\frac{1}{49x50}\)

= \(\frac{1}{2}\)-  \(\frac{1}{3}\)+  \(\frac{1}{3}\)-  \(\frac{1}{4}\)+  ...  +  \(\frac{1}{49}\)-  \(\frac{1}{50}\)

=  \(\frac{1}{2}\)-  \(\frac{1}{50}\)

=\(\frac{12}{25}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\frac{1}{2}-\frac{1}{50}\)

=\(=\frac{12}{15}\)

Vãi cả nhân :V

\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+\frac{2}{5\cdot6}\\ =2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\\ =2\left(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+\frac{5-4}{4\cdot5}+\frac{6-5}{5\cdot6}\right)\\ =2\left(\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+\frac{5}{4\cdot5}-\frac{4}{4\cdot5}+\frac{6}{5\cdot6}-\frac{5}{5\cdot6}\right)\\ =2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{6}\right)\\ =2\left(1-\frac{1}{6}\right)\\ =2\cdot\frac{5}{6}=\frac{10}{6}\)

Chúc bạn học tốt nha.

hình như là 6666666.6

Ta có:

\(\frac{1}{3.4}.x+\frac{1}{4.5}.x+...+\frac{1}{49.50}.x=1\)

\(\Rightarrow x.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)=1\)

\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)

\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{50}\right)=1\Leftrightarrow x.\frac{47}{150}=1\)

\(\Rightarrow x=1:\frac{47}{150}\Leftrightarrow x=\frac{150}{47}\)

\(a.\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)

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

\(=\frac{3}{10}\)

\(b.\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}\)

\(=2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}\right)\)

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

\(=2\cdot\left(\frac{1}{2}-\frac{1}{5}\right)\)

\(=2\cdot\frac{3}{10}=\frac{3}{5}\)

\(c.\frac{1}{2\cdot3}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}\)

\(=\frac{1}{6}+\frac{2}{15}+\frac{3}{40}\)

\(=\frac{3}{8}\)

k nha 500 AE

a, \(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}\)

\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}\)

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

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

\(=\frac{3}{10}\)

b, \(\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}\)

\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}\)

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

\(=\left(\frac{1}{2}-\frac{1}{5}\right)\times\frac{2}{1}\)

\(=\frac{3}{10}\times\frac{2}{1}\)

\(=\frac{3}{5}\)

c,  \(\frac{1}{2\times3}+\frac{2}{3\times5}+\frac{3}{5\times8}\)

 \(=\frac{3-2}{2\times3}+\frac{5-3}{3\times5}+\frac{8-5}{5\times8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}\)

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

\(=\frac{3}{8}\)

Bạn Nguyễn Đoàn Tâm chỉ đúng câu a và b thôi còn câu c sai rùi . sorry, mình nhìn nhầm

\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

\(=\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+\frac{5-4}{4\times5}+\frac{6-5}{5\times6}\)

\(=\frac{3}{2\times3}-\frac{2}{2\times3}+\frac{4}{3\times4}-\frac{3}{3\times4}+\frac{5}{4\times5}-\frac{4}{4\times5}+\frac{6}{5\times6}-\frac{5}{5\times6}\)

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

\(=\frac{1}{2}-\frac{1}{6}\)

\(=\frac{1}{3}\)

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

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

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

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

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

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

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

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

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

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

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

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

\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)

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

=\(1-\frac{1}{9}\)

=\(\frac{8}{9}\)

OK XONG NHỚ CHO MIK NHA

\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)

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

=1-\(\frac{1}{9}\)

=\(\frac{8}{9}\)

\(\frac{1}{1\times2}+........+\frac{1}{8\times9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{9}-\frac{1}{10}\)

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