\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)

\(M=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)

\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)

\(=\frac{1}{2}-\frac{1}{11.12}\)

 \(=\frac{65}{132}\)

65/132

65/132

\(2M=2\cdot\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{10\cdot11\cdot12}\right)\)

\(2M=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+.....+\frac{2}{10\cdot11\cdot12}\)

\(2M=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+.....+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\)

\(2M=\frac{1}{1\cdot2}-\frac{1}{11\cdot12}\)

\(2M=\frac{1}{2}-\frac{1}{132}\)

\(2M=\frac{66}{132}-\frac{1}{132}\)

\(2M=\frac{65}{132}\)

\(M=\frac{65}{132}:2\)

\(M=\frac{65}{264}\)

Giải:

Ta có nhận xét:

\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{3-1}{1.2.3}=\frac{2}{1.2.3}\)

\(\frac{1}{2.3}-\frac{1}{3.4}=\frac{4-2}{2.3.4}=\frac{2}{2.3.4}\)

=>\(\frac{1}{1.2.3}=\frac{1}{3}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)\)

\(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)\)

Do đó M=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}-\frac{1}{11.12}\right)\)

=\(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)=\frac{1}{2}-\frac{1}{11.12}\)

=\(\frac{1}{2}.\frac{65}{132}=\frac{65}{124}\)

Vậy M=65/124

M=\(\frac{65}{124}\)

65/124 nha bạn

M= 1 - 1/2 + 1/2 - 1/3 +1/3 -1 /4 + 1/4 - 1/5 + .....+ 1/10 - 1/11 + 1/11 -1/12​

   = 1 - 1/12

   = 11/12

=\(\frac{65}{264}\)

65/264 k cho mình nhé

Ta có : \(\frac{1}{1\cdot2\cdot3}=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)\)

            \(\frac{1}{2\cdot3\cdot4}=\frac{1}{2}\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)\)

                    ...

            \(\frac{1}{10\cdot11\cdot12}=\frac{1}{2}\left(\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)

\(\Rightarrow y=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2\cdot3}-\frac{1}{3.4}\right)+.........+\frac{1}{2}\left(\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)

\(\Rightarrow y=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)

\(\Rightarrow y=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{11\cdot12}\right)\)

\(\Rightarrow y=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{132}\right)\Rightarrow y=\frac{1}{2}\cdot\frac{65}{132}\Rightarrow y=\frac{65}{264}\)

Đặt biểu thức là A

\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+.2018.2019\)

\(2A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)

\(2A=\frac{1}{2}-\frac{1}{2018.2019}\)

A= 1/4 - 1/(2018.2019)

Vậy A = ... (tự ghi)

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2017.2018.2019}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2017.2018.2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2018.2019}\right)\)

Tự làm nốt

\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2017.2018.2019}\)

\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2017.2018.2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2018.2019}\right)\)

Đến đây tự làm tiếp :))) 

M = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/10.11.12

M = 1/2.(2/1.2.3 + 2/2.3.4 + 2/3.4.5 + ... + 2/10.11.12)

M = 1/2.(1/1.2 - 1/2.3 + 1/2.3- 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/10.11 - 1/11.12)

M = 1/2.(1/1.2 - 1/11.12)

M = 1/4 - 1/264

M = 65/264

M=65/264.

Kết quả 65/264

   -chúc các bạn học tốt-

Xét ct trước :D

\(\frac{2}{\left[\left(n-1\right)n\left(n+1\right)\right]}=\frac{1}{\left[\left(n-1\right)n\right]}-\frac{1}{\left[n\left(n+1\right)\right]}\)

Sau khi xét ct rồi thì /Bùm/ Ta được: 

\(2M=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{10.11.12}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)

\(=\frac{1}{1.2}-\frac{1}{11.12}\)

\(=\frac{65}{132}\)

\(\Rightarrow M=\frac{65}{264}\)

Ok rồi nhé :)