I. Tìm x,biết :
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right)2x}=\frac{11}{48}\)
( \(x\in N,x\ge2)\)
K
Khách
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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
1 tháng 6 2018
Bài 1:
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(x-1\right).x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x-1}-\frac{1}{x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4.}.\left(1-\frac{1}{x}\right)=\frac{11}{48}\)
\(1-\frac{1}{x}=\frac{11}{48}:\frac{1}{4}\)
\(1-\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{x}=1-\frac{11}{12}\)
\(\frac{1}{x}=\frac{1}{12}\)
Vậy x= 12
Bài 2 :
Xét vế trái ta có :
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{\left(3n-1\right).\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{1}{2\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)
VẾ TRÁI ĐÚNG BẰNG VẾ PHẢI .ĐẲNG THỨC ĐÃ CHỨNG TỎ LÀ ĐÚNG
cHÚC BẠN HỌC TỐT ( -_- )
16 tháng 6 2020
\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)\cdot2x}=\frac{1}{8}\left(x\inℕ;x\ge2\right)\)
Đặt \(A=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)2x}\)
\(2A=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left(2x-2\right)2x}\)
\(2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2x-2}-\frac{1}{2x}\)
\(2A=\frac{1}{2}-\frac{1}{2x}=\frac{x-1}{2x}\)
\(\Rightarrow A=\frac{x-1}{2x}:2=\frac{x-1}{2x}\cdot\frac{1}{2}=\frac{x-1}{4x}\)
Mà \(A=\frac{1}{8}\Rightarrow\frac{x-1}{4}=\frac{1}{8}\)
\(\Leftrightarrow8x-8=4\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\frac{12}{8}=\frac{3}{2}\left(ktm\right)\)
Vậy không có x thỏa mãn yêu cầu đề bài
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
\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Rightarrow2x=24\)
\(\Rightarrow x=12\)
LB
4 tháng 5 2018
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right)2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Rightarrow x=12\) (nh)
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)
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
*Bài làm:
~I) Tìm x:
➤Ta có: \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + ... + \(\frac{1}{\left(2x-2\right)2x}\) = \(\frac{11}{48}\)
⇒ \(2\) . (\(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + ... + \(\frac{1}{\left(2x-2\right)2x}\)) = \(2\) . \(\frac{11}{48}\)
⇒ \(\frac{2}{2.4}\) + \(\frac{2}{4.6}\) + ... + \(\frac{2}{\left(2x-2\right)2x}\) = \(\frac{22}{48}\)
⇒ (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + (\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + (\(\frac{1}{2x-2}\) - \(\frac{1}{2x}\)) = \(\frac{22}{48}\)
⇒ \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + \(\frac{1}{6}\) - ... - \(\frac{1}{2x-2}\) + \(\frac{1}{2x-2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{1}{2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x}{x}\) . \(\frac{1}{2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x}{2x}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x-1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x-1}{2x}\) = \(\frac{22}{48}\)
⇒ \(x-1\) = \(\frac{22}{48}\) . \(2x\)
⇒ \(x-1\) = \(\frac{44x}{48}\)
⇒ \(x\) = \(\frac{44x}{48}\) + \(1\)
⇒ \(x\) = \(\frac{44x}{48}\) + \(\frac{48}{48}\)
⇒ \(x\) = \(\frac{44x+48}{48}\)
⇒ \(x\) = \(12\) (Chỗ này mình bấm máy tính nên hơi tắt;Bạn thông cảm)
*Vậy \(x\) = \(12\) .