Tính tổng:
D=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)có lời giải nhé
tính
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+..............+\frac{1}{10.11.12}\)
\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)
TÍnh giá trị biểu thức :
M = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)
\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{11.12}\)
\(=\frac{65}{132}\)
Tính M biết:
M = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)
Bạn nào giải nhanh hộ mk với đg gấp ^_^
\(2M=2\cdot\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{10\cdot11\cdot12}\right)\)
\(2M=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+.....+\frac{2}{10\cdot11\cdot12}\)
\(2M=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+.....+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\)
\(2M=\frac{1}{1\cdot2}-\frac{1}{11\cdot12}\)
\(2M=\frac{1}{2}-\frac{1}{132}\)
\(2M=\frac{66}{132}-\frac{1}{132}\)
\(2M=\frac{65}{132}\)
\(M=\frac{65}{132}:2\)
\(M=\frac{65}{264}\)
Tính giá trị biểu thức:
M=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{10.11.12}\)
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=\(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+ \(\frac{1}{3.4.5}\)+......... + \(\frac{1}{10.11.12}\)
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
\(y=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.........+\frac{1}{10.11.12}\)
Lam giup minh voi nhanh len minh sap nop roi!!!!!!!!!!!!!!!!
Ta có : \(\frac{1}{1\cdot2\cdot3}=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)\)
\(\frac{1}{2\cdot3\cdot4}=\frac{1}{2}\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)\)
...
\(\frac{1}{10\cdot11\cdot12}=\frac{1}{2}\left(\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)
\(\Rightarrow y=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2\cdot3}-\frac{1}{3.4}\right)+.........+\frac{1}{2}\left(\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)
\(\Rightarrow y=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)
\(\Rightarrow y=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{11\cdot12}\right)\)
\(\Rightarrow y=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{132}\right)\Rightarrow y=\frac{1}{2}\cdot\frac{65}{132}\Rightarrow y=\frac{65}{264}\)
Tính:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2017.2018.2019}\)
Giải chi tiết giúp mình nhé
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 :)))
Tính giá trị biểu thức M=\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) +............\(\frac{1}{10.11.12}\)
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
Tính giá trị biểu thức :
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)
Ai làm đúng 3 tik
T biết làm nhưng vẫn đăng để mấy pẹn trl
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 :)