Tính nhanh:
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\)
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AH
Akai Haruma
Giáo viên
31 tháng 5 2023
Lời giải:
Gọi tổng trong ngoặc là $A$
$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{10-8}{8.9.10}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}$
$=\frac{1}{1.2}-\frac{1}{9.10}=\frac{1}{2}-\frac{1}{90}=\frac{22}{45}$
Vậy $\frac{22}{45}x=\frac{23}{45}$
$\Rightarrow x=\frac{23}{45}: \frac{22}{45}=\frac{23}{22}$
AH
Akai Haruma
Giáo viên
31 tháng 3 2023
$x$ ở cuối là sao đây bạn? Nhân riêng với $\frac{1}{8.9.10}$ à?
SN
10 tháng 8 2021
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-...-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right).x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.\dfrac{22}{45}.x=\dfrac{22}{45}\)
=> \(\dfrac{1}{2}.x=1\)
=> \(x=2\)
Vậy x = 2
Chúc bạn học tốt !!!
19 tháng 7 2017
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{23}{45}\)
\(\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\right].x=\dfrac{23}{45}\)\(\left[\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)\right].x=\dfrac{23}{45}\)
\(\left(\dfrac{1}{2}.\dfrac{22}{45}\right).x=\dfrac{23}{45}\)
\(\dfrac{11}{45}.x=\dfrac{23}{45}\)
\(x=\dfrac{23}{45}:\dfrac{11}{45}\)
\(x=\dfrac{23}{11}\)
17 tháng 5 2017
Ta có:
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right)x=\dfrac{23}{45}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{8.9.10}\right)x\) \(=\dfrac{23}{45}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)x\) \(=\dfrac{23}{45}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)x=\dfrac{23}{45}\)
\(\Leftrightarrow\dfrac{11}{45}.x=\dfrac{23}{45}\Leftrightarrow x=\dfrac{23}{45}\div\dfrac{11}{45}=\dfrac{23}{11}\)
Vậy \(x=\dfrac{23}{11}\)
1
OY
26 tháng 9 2021
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
NL
1
17 tháng 5 2022
\(2S=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{23+24+25}=\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+\left(\dfrac{1}{23.24}-\dfrac{1}{24.25}\right)\)\(=\dfrac{1}{1.2}-\dfrac{1}{24.25}=\dfrac{299}{600}\)
Vậy \(S=\dfrac{299}{600}\div2=\dfrac{299}{1200}\)
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\)
\(S=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(S=\dfrac{1}{1.2}-\dfrac{1}{9.10}\)
\(S=\dfrac{1}{2}-\dfrac{1}{90}=\dfrac{44}{90}\)