Tìm x biết :
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)
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HP
25 tháng 2 2016
=>(2/1.2.3+2/2.3.4+....+2/8.9.10).x=22/45
=>(1/1.2-1/2.3+1/2.3-1/3.4+....+1/8.9-1/9.10).x=22/45
=>(1/1.2-1/9.10).x=22/45
=>22/45.x=44/45
=>x=2
HP
25 tháng 2 2016
4+2^2+2^3+....+2^20=2^n
=>2^2+2^2+2^3+....+2^20=2^n
đặt 2^2+2^3+....+2^20
=>2A-A=2^21-2^2
khi đó A=2^2+2^21-2^2=2^21=2^n
=>n=21
31 tháng 1 2016
Đặt B = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{8.9.10}\)
=> 2B = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{8.9.10}\)
=> 2B = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\)
=> 2B = \(\frac{1}{1.2}-\frac{1}{9.10}\)
2B = \(\frac{22}{45}\)
B = \(\frac{22}{45}:2\)
=> B = \(\frac{11}{45}\)
Ta có : \(\frac{11}{45}.x=\frac{22}{45}\)
=> x = \(\frac{22}{45}:\frac{11}{45}\)
=> x = \(\frac{2}{1}\)
19 tháng 2 2017
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{22}{45}x=\frac{44}{45}\)
\(\Rightarrow x=\frac{44}{45}:\frac{22}{45}\)
\(\Rightarrow x=2\)
Vậy x = 2
19 tháng 2 2017
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2=\frac{22}{45}:x\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2=\frac{22}{45}:x\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right):2=\frac{22}{45}:x\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2=\frac{22}{45}:x\)
\(\Rightarrow\frac{22}{45}:2=\frac{22}{45}:x\)
\(\Rightarrow x=2\)
AK
9 tháng 4 2018
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{45}{90}-\frac{1}{90}\right).x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{44}{90}.x=\frac{22}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{22}{45}\)
\(\Rightarrow x=\frac{22}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{22}{45}.\frac{45}{11}\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
Chúc học tốt !!!
AK
9 tháng 4 2018
Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\frac{2}{6}=\frac{1}{6}=\frac{1}{1.2.3}\)
VD ( dễ hiểu )
D
12 tháng 10 2019
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}\cdot\frac{22}{45}=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}=1\)
\(\Rightarrow x=2\)
5 tháng 12 2015
\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\frac{22}{45}.x=\frac{44}{45}\)
x =2
HP
21 tháng 2 2016
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)
\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\Leftrightarrow\frac{22}{45}.x=\frac{44}{45}\Leftrightarrow x=2\)
Vậy x=2
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
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right).x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
\(pt\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Leftrightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)
\(\Leftrightarrow\frac{1}{2}x=1\)
\(\Rightarrow x=2\)