Tìm x, biết:
1/2.4+1/4.6+...+1/(2x-2).2.x Giúp mk nha
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Khách
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HV
1
26 tháng 8 2021
Giải
1/(2.4) + 1/(4.6) + … + 1/[(2x – 2).2x] = 1/8
=> 2/(2.4) + 2/(4.6) + ...+ 2/[(2x - 2).2x] = 2/8
=>1-1/4+1/4-1/6+...+1/(2x-2) - 1/2x = 2/8
=>1 - 1/2x = 2/8
=>1/2x = 1 - 2/8
=>1/2x = 6/8 = 3/4
=>1.4 = 2.x.3
=>4 = 6x
=> x thuộc rỗng
Vậy x thuộc rỗng
22 tháng 4 2018
1/(2.4) + 1/(4.6) + … + 1/[(2x – 2).2x] = 1/8
suy ra 2/(2.4) + 2/(4.6) + ...+ 2/[(2x - 2).2x] = 2/8
suy ra 1-1/4+1/4-1/6+...+1/(2x-2) - 1/2x = 2/8
suy ra 1 - 1/2x = 2/8
suy ra 1/2x = 1 - 2/8
suy ra 1/2x = 6/8 = 3/4
suy ra 1.4 = 2.x.3
suy ra 4 = 6x
suy ra x thuộc rỗng
Vậy x thuộc rỗng
k cho mình nha. Chúc bạn học tốt!
NA
2
4 tháng 5 2018
\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)\cdot2x}=\frac{1}{8}\)
\(\Rightarrow\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left(2x-2\right)\cdot2x}=\frac{2}{8}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
6 tháng 5 2018
=>\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)=\frac{1}{8}\)
=>\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)
=>\(\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
=>\(\frac{1}{2x}=\frac{1}{4}\)
=> \(2x=4\)
=> \(x=2\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2023
\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Rightarrow2x=24\)
\(\Rightarrow x=12\)
KT
6 tháng 2 2018
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{\left(2x-2\right).2x}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2x}\right)\)
\(=2.\frac{x-1}{2x}\)
\(=\frac{x-1}{x}\)
đề thiếu
LT
2
30 tháng 4 2018
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)= \(\frac{1}{8}\)
\(\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}\right)\)= \(\frac{1}{8}\)
\(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2x}\right)\)= \(\frac{1}{8}\)
=> \(\frac{1}{2}-\frac{1}{2x}\)= \(\frac{1}{4}\)
=> 1/2x = 1/4
=> 2x = 4
x = 4 : 2
x = 2
30 tháng 4 2018
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right)2x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{8}.2=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)
\(\Rightarrow2x=4\Leftrightarrow x=2\)
23 tháng 2 2022
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)
DT
23 tháng 2 2022
\(\Leftrightarrow\dfrac{1}{4}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.......+\dfrac{1}{\left(x-1\right)x}\right)=\dfrac{1}{8}\) ( đk x khác 0 , x khác 1)
\(\Leftrightarrow\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{x-1}-\dfrac{1}{x}\right)=\dfrac{1}{8}\)
\(\Leftrightarrow1-\dfrac{1}{x}=\dfrac{1}{2}\)
=> x =2 ( tm)
KN
1 tháng 5 2019
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}\right)=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{8}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
26 tháng 6 2019
TL:
\(\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{\left(2x-2\right)2x}\right)=\frac{1}{8}\)
\(\frac{1}{2}-\frac{1}{4x}=\frac{1}{8}\)
\(\frac{1}{4x}=\frac{3}{8}\)
=>x=2/3
hc tốt
\(\text{Sửa đề:}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(\text{Đặt biểu thức là A:}\)
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}=\frac{1}{8}\times2=\frac{1}{4}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{4}\)
\(2A=\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(A=\frac{1}{2}\times\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)
\(A=\frac{1}{2}\times\frac{1}{2}-\frac{1}{2}\times\frac{1}{2x}=\frac{1}{8}\)
\(A=\frac{1}{4}-\frac{1}{4x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{4x}=\frac{1}{4}-\frac{1}{8}=\frac{2}{8}-\frac{1}{8}=\frac{1}{8}\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=8\div4=2\)