\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)

\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)

:V Làm sai hết rồi sai ngay từ bước đầu tiên.

\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)

\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)

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

\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(=\frac{1}{12}-\frac{3}{20}\)

\(=\frac{-11}{12}\)

\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)

= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

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

= \(-\frac{7}{30}\)

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

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

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

~ Hok tốt ~

1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10

=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

=1/2-1/10

=5/10-1/10

=4/10=2/5

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)

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

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

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

= 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

= 1/2-0-0-0-0-0-0-0-0-1/10

=1/2 - 1/10 

=2/5

tớ trước nhé

a) \(5\frac{8}{17}:x+\frac{-1}{17}:x+3\frac{1}{17}:17\frac{1}{3}=\frac{4}{17}\)

\(\frac{93}{17}:x+\frac{-1}{17}:x+\frac{52}{17}:\frac{52}{3}=\frac{4}{17}\)

\(\left(\frac{93}{17}+\frac{-1}{17}\right):x+\frac{52}{17}.\frac{3}{52}=\frac{4}{17}\)

\(\frac{92}{17}:x+\frac{3}{17}=\frac{4}{17}\)

\(\frac{92}{17}:x=\frac{4}{17}-\frac{3}{17}\)

\(\frac{92}{17}:x=\frac{1}{17}\)

\(x=\frac{92}{17}:\frac{1}{17}\)

\(x=92\)

b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{1}{3}.\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)

\(1-\frac{1}{x+3}=\frac{18}{19}\)

\(\frac{1}{x+3}=1-\frac{18}{19}\)

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

\(\Rightarrow x+3=19\)

\(\Rightarrow x=19-3\)

\(\Rightarrow x=16\)

Ta có:

\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)

\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)

\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)

\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)

A.  = 1/2-1/3+1/3-1/4+1/4-1/5...+1/101-1/102=1/2-1/102=25/51.

B.  =1/5-1/10+1/10-1/15+...+1/115-1/120=1/5-1/120=23/120.

C.  = 1/5-1/7+1/7-1/9+1/9-1/11+...+1/997-1/999=1/5-1/999=994/4995.

Minh kiem tra bang may tinh roi do.

\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{101\times102}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}\)

\(=1-\frac{1}{102}\)

\(=\frac{101}{102}\)

\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}\cdot\frac{17}{4}-28\cdot\frac{4}{3}\right):\frac{7}{4}\)

\(=\frac{59}{15}-\frac{29}{4}:\frac{7}{4}=\)\(\frac{59}{15}-\frac{29}{7}=\frac{-22}{105}\)

B = \(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{17}{4}-2x\frac{4}{3}\right):\frac{7}{4}\)

    = \(\frac{59}{10}x\frac{2}{3}-\left(\frac{119}{12}-\frac{8}{3}\right)x\frac{4}{7}\)

    = \(\frac{59}{15}-\frac{29}{4}x\frac{4}{7}=\frac{59}{15}-\frac{29}{7}\)

     = \(\frac{-22}{105}\)

C = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)

    = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

     = \(1-\frac{1}{7}=\frac{6}{7}\)

\(B=\frac{59}{15}-\frac{29}{7}\)

 \(=\frac{-22}{105}\) ( bài B này mk hk pít giải nhanh mk bấm máy tính lụi)

\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\)\(\left(vì\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\right)\)

\(C=1-\frac{1}{7}=\frac{6}{7}\)

1=3/3=4/4=5/5=...

=> 1+1/1*3=3/1*3=1/1

=> 1+1/2*4=4/2*4=1/2

=>...

Bieu thuc se con lai la 1*1/2*1/3*1/4*1/5

Vay A=1/120