\(\frac{1}{1.2.3}\)+ \(\frac{1}{2.3.4}\)+....+ \(\frac{1}{7.8.9}\)+ \(\frac{1}{8.9.10}\)
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VH
1 tháng 3 2016
1/2x3x4 + 1/3x4x5 + 1/4x5x6 + 1/5x6x7 + ..... + 1/8x9x10
= { 2/2x3x4 + 2/3x4x5 + 2/4x5x6 + .... + 2/8x9x10 } : 2
= { 4-2/2x3x4 + 5-3/3x4x5 + 6-4/4x5x6 + .... + 10-8/8x9x10 } : 2
= { 4/2x3x4 - 2/2x3x4 + 5/3x4x5 - 3/3x4x5 + ... + 10/8x9x10 - 8/8x9x10 } : 2
= { 1/2x3 - 1/3x4 + 1/3x4 - 1/4x5 + ... + 1/8x9 - 1/9x10 } : 2
= { 1/2x3 - 1/9x10 } :2
= { 1/6 - 1/90 } : 2
= 14/90 : 2
= 7/90
21 tháng 5 2018
Bài 1 \(F=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{7.8.9}+\frac{1}{8.9.10}\)
\(2F=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{7.8}-\frac{1}{8.9}+\frac{1}{8.9}-\frac{1}{9.10}\)
\(2F=\frac{1}{1.2}-\frac{1}{9.10}\)\(=\frac{44}{90}\)
\(F=\frac{11}{45}\)
Vậy \(F=\frac{11}{45}\)
Bài 2 :
\(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\Rightarrow\)\(\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)
\(\Rightarrow\)\(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}< B< \frac{1}{2.3}+..+\frac{1}{8.9}\)
\(\Rightarrow\)\(\frac{1}{3}-\frac{1}{10}< B< \frac{1}{2}-\frac{1}{9}\)
\(\Rightarrow\)\(\frac{7}{30}\)\(< \frac{7}{18}\left(đpcm\right)\)
Hết nha bn.Mk ik ngủ.Chúc bạn học tốt
DL
23 tháng 2 2020
Có \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)+x=\frac{23}{45}\)
Cho \(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\)
Ta có công thức sau: \(\frac{1}{n\cdot\left(n+1\right)}+\frac{1}{\left(n+1\right)\cdot\left(n+2\right)}=\frac{2}{n\cdot\left(n+1\right)\left(n+1\right)}\)
\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\\ =\frac{1}{1\cdot2}-\frac{1}{9\cdot10}=\frac{22}{45}\)
\(\Rightarrow A=\frac{22}{45}:2=\frac{11}{45}\)
Thay vào phép tính trên ta được:
\(\frac{11}{45}\cdot x=\frac{23}{45}\\ x=\frac{23}{45}:\frac{11}{45}\\ x=\frac{23}{11}\)
Vậy \(x=\frac{23}{11}\)
12 tháng 1 2017
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
5 tháng 4 2017
đặt \(A=\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)
\(A=\frac{22}{45}:2=\frac{11}{45}\)
thay A vào ta được
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)
XO
13 tháng 8 2019
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
22 tháng 3 2016
Ta có:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right)\)
=\(\frac{11}{45}\)
22 tháng 3 2016
=1/2.(1/1*2-1/2*3+1/2*3-1/3*4+...+1/8*9-1/9*10)
=1/2.(1/1*2-1/9*10)
=1/2.44/90
=22/90
nha
7 tháng 6 2020
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{23}{45}\)
=> \(\left[\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\right)\right]x=\frac{23}{45}\)
=>\(\left[\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)
=> \(\left[\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9\cdot10}\right)\right]x=\frac{23}{45}\)
=> \(\left[\frac{1}{2}\cdot\frac{22}{45}\right]x=\frac{23}{45}\)
=> \(\frac{11}{45}x=\frac{23}{45}\)
=> \(x=\frac{23}{45}:\frac{11}{45}=\frac{23}{45}\cdot\frac{45}{11}=\frac{23}{11}\)
Vậy x = 23/11
Ez :))
NT
1
16 tháng 3 2017
\(pt\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Leftrightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)
\(\Leftrightarrow\frac{1}{2}x=1\)
\(\Rightarrow x=2\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+..........+\frac{1}{8.9}-\frac{1}{9.10}\)
\(=\frac{1}{1.2}-\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{90}\)
\(=\frac{45}{90}-\frac{1}{90}\)
\(=\frac{44}{90}\)
\(=\frac{22}{45}\)
22/45