\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{11}{48}\left(x\in N;x\ge12\right)\)
K
Khách
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8 tháng 3 2017
\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2x-2\right)2x}=\dfrac{11}{48}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2x-2\right)2x}\right)=\dfrac{11}{48}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{11}{48}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Leftrightarrow2x=24\Leftrightarrow x=12\) (thỏa mãn)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2023
\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Rightarrow2x=24\)
\(\Rightarrow x=12\)
NT
10 tháng 5 2018
mình ko biết mình làm đúng hay sai bạn nhé, mong mọi người góp ý
= 1/2.( 1/2.4+1/4.6+....+1/(2x-2)2x)=1/8
= 1/2.(1/2-1/4+1/4-1/6+....+1/(2x-2)-1/2x)=1/8
= 1/2.( 1/2-1/2x)=1/8
( 1/2-1/2x)=1/8:1/2
1/2-1/2x=1/4
1/2x =1/2-1/4
1/2x =1/4
2x = 4
x =4:2
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)
LB
4 tháng 5 2018
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-1}{2x}=\dfrac{-1}{4}\)
\(\Rightarrow x=2\)
S
4 tháng 5 2018
Ta có: \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\Rightarrow\dfrac{1}{2}.\dfrac{x-1}{2x}=\dfrac{1}{8}\Rightarrow\dfrac{x-1}{4x}=\dfrac{1}{8}\)
\(\Rightarrow8\left(x-1\right)=4x\Rightarrow8x-8=4x\Rightarrow4x=8\Rightarrow x=2\)
2 tháng 11 2023
Gọi biểu thức trên là A
Ta có:
2A = (\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+...+\(\dfrac{1}{x.\left(x+2\right)}\)).2
2A = \(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+...+\(\dfrac{2}{x\left(x+2\right)}\)
2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\)
2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{x+2}\)
mà A = \(\dfrac{1}{10}\)(đề bài)
nên 2A = \(\dfrac{2}{10}\) hay \(\dfrac{1}{2}\) - \(\dfrac{1}{x+2}\) = \(\dfrac{2}{10}\)
suy ra \(\dfrac{1}{x+2}\) = \(\dfrac{1}{2}\)-\(\dfrac{2}{10}\)=\(\dfrac{3}{10}\)
AH
Akai Haruma
Giáo viên
8 tháng 3 2021
Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 3 2023
\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)
\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right)2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Rightarrow x=12\) (nh)