Tìm số tự nhiên x biết :
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{23}{45}\)
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19 tháng 7 2017
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{23}{45}\)
\(\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\right].x=\dfrac{23}{45}\)\(\left[\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)\right].x=\dfrac{23}{45}\)
\(\left(\dfrac{1}{2}.\dfrac{22}{45}\right).x=\dfrac{23}{45}\)
\(\dfrac{11}{45}.x=\dfrac{23}{45}\)
\(x=\dfrac{23}{45}:\dfrac{11}{45}\)
\(x=\dfrac{23}{11}\)
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 $\frac{22}{45}x=\frac{23}{45}$
$\Rightarrow x=\frac{23}{45}: \frac{22}{45}=\frac{23}{22}$
20 tháng 7 2017
1.
\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...........+\dfrac{1}{8.9.10}\right)x=\dfrac{23}{45}\)
\(\Leftrightarrow\left[\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+............+\dfrac{2}{8.9.10}\right)\right]x=\dfrac{23}{45}\)
\(\Leftrightarrow\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+........+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\right]x=\dfrac{23}{45}\)
\(\Leftrightarrow\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{8.9}\right)\right]x=\dfrac{23}{45}\)
\(\Leftrightarrow\left[\dfrac{1}{2}.\dfrac{22}{45}\right]x=\dfrac{23}{45}\)
\(\Leftrightarrow\dfrac{11}{45}.x=\dfrac{23}{45}\)
\(\Leftrightarrow x=\dfrac{23}{11}\)
Vậy \(x=\dfrac{23}{11}\) là giá trị cần tìm
2.
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.........+\dfrac{1}{x\left(x+1\right):2}=\dfrac{1998}{2000}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...............+\dfrac{2}{x\left(x+1\right)}=\dfrac{1998}{2000}\)
\(\Leftrightarrow\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...........+\dfrac{2}{x\left(x+1\right)}=\dfrac{1998}{2000}\)
\(\Leftrightarrow2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+.........+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1998}{2000}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.........+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{999}{2000}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2000}\)
\(\Leftrightarrow x+1=2000\)
\(\Leftrightarrow x=1999\)
Vậy \(x=1999\) là giá trị cần tìm
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 !!!
D
22 tháng 5 2017
Ta có:
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)
Vậy \(x=\frac{23}{11}\)
DP
23 tháng 5 2017
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{8.9}+\frac{1}{9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{72}+\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\left(\frac{1}{2}-\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\frac{22}{45}.\frac{1}{2}x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}\div\frac{11}{45}\)
\(x=\frac{23}{11}\)
=> \(x=\frac{23}{11}\)
2 tháng 8 2015
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{46}{45}\)
\(=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right).x\)
\(=\left(\frac{1}{2}-\frac{1}{90}\right).x=\left(\frac{45}{90}-\frac{1}{90}\right)x=\frac{44}{90}.x=\frac{22x}{45}=\frac{46}{45}\)
=> 22x=46
=> x=\(46:22=\frac{23}{11}\)
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 \(x=\dfrac{23}{11}\)