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

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

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

Ở link này có bài tham khảo nè bn :

http://olm.vn/hoi-dap/detail/42438427638.html

Zoro  cảm ơn nhiều

Với \(a\inℕ;a>0\), ta có:

\(\frac{2}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2-a}{a\left(a+1\right)\left(a+2\right)}=\frac{a+2}{a\left(a+1\right)\left(a+2\right)}-\frac{a}{a\left(a+1\right)\left(a+2\right)}\)

\(=\frac{1}{a\left(a+1\right)}-\frac{1}{\left(a+1\right)\left(a+2\right)}\)

Theo đề :

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{44}{45}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{44}{45}\Leftrightarrow\frac{44}{90}x=\frac{44}{45}\Leftrightarrow x=\frac{44}{45}:\frac{44}{90}=\frac{90}{45}=2\)

Vậy \(x=2\)

Chúc bạn học tốt!

P/S: đây là 1 trong những dạng bài toán hay của lớp 6: tính toán trên dãy số. Để làm được dạng này thì mình có 2 quy tắc tổng quát như sau:

1)Khử liên tiếp

2)Đưa về những dãy số cơ bản đã biết cách tính (VD: 1+2+3+...; ...)

= \(\frac{1x1x1}{1x2x4}x\frac{2.2.1}{1.1.2.2}=\frac{1}{8}.1=\frac{1}{8}\)

=1X2X3/1X2X3X4X2= 1/8                 =3X2X2X2X5/3X2X2X5X2= 1/1

=1/8X1/1=1/8

Tính giá trị biểu thức:

\(\frac{1\times2\times3}{1\times6\times8}\times\frac{6\times4\times5}{3\times2\times2\times10}\)   

\(=\frac{1}{8}\times1=\frac{1}{8}\)

Ủng hộ nha! :) 

\(=\frac{47}{6}\)

Ta thấy :

1/1x2 = 1/1 - 1/2

1/2x3 = 1/2 - 1/3 

....

=>( 1/1x2 + 1/2x3 + 1/3x4 + 1/5x6 ) x 10 - x = ( 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 ) x 10 - x

= ( 1/1 - 1/6 ) x 10 - x =0

 5/6 x 10 - x = 0 

 25/3 - x = 0

=> x = 25/3

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{5.6}\right).10-x=0\)

\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{30}\right).10-x=0\)

\(\left(1-\frac{1}{4}+\frac{1}{30}\right).10-x=0\)

\(\left(\frac{3}{4}+\frac{1}{30}\right).10-x=0\)

\(\frac{47}{60}.10-x=0\)

\(\frac{47}{6}-x=0\)

\(x=\frac{47}{6}-0\)

\(x=\frac{47}{6}\)

B=2+1/1.2+2+1/2.3+........+2+1/9.10

B=2.9+1/1.2+1/2.3+........+1/9.10

B=18+9/10

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

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

=\(2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\right)\)

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

=\(2\times\left(\frac{1}{1}-\frac{1}{10}\right)=2\times\left(\frac{10}{10}-\frac{1}{10}\right)=2\times\frac{9}{10}\)

=\(\frac{9}{5}\)

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

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

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

=\(\frac{9}{5}\).

**** mình nha mấy bạn.

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

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

\(\frac{2}{5}x=\frac{9}{10}-\frac{3}{10}=\frac{3}{5}\)

\(x=\frac{\frac{3}{5}}{\frac{2}{5}}=\frac{3}{2}\)

Ta có: \(\frac{1}{1x2}\)+ \(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ .....+ \(\frac{1}{9x10}\)

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

        = 1 - \(\frac{1}{10}\)

        =  \(\frac{9}{10}\)

X = \(\frac{3}{2}\) là đúng đó bạn

=99/100       

=> \(C=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

C = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)

C = \(1-\frac{1}{100}