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Khách

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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

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

11 tháng 8 2019

*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 \(x\) = \(12\) .

11 tháng 8 2019

Violympic toán 7

6 tháng 5 2018

=>\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right)=\frac{1}{8}\)

=>\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2x}\right)=\frac{1}{8}\)

=>\(\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)

=>\(\frac{1}{2x}=\frac{1}{4}\)

=> \(2x=4\)

=> \(x=2\)

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 ( -_- )

28 tháng 4 2018

\(\Rightarrow2.\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{\left(2x-2\right).2x}\right)=\frac{1}{8}.2\).2

\(\Rightarrow\frac{2}{2.4}+\frac{2}{4.6}+...\frac{2}{\left(2x-2\right).2x}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\Rightarrow\frac{1}{2x}=\frac{1}{2.2}\)

\(\Rightarrow x=2\)

28 tháng 4 2018

Đề có sai ko bn ?

10 tháng 5 2018

Cần gấp, mai thi

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

5 tháng 4 2015

Gọi biều thức trên là A, ta có:

A=(1/2.4+1/4.6+1/6.8+1/8.10+1/10.12)x=2

2A=(2/2.4+2/4.6+2/6.8+2/8.10+2/10.12)x=2

2A=(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)x=2

2A=(1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12)x=2

2A=(1/2-1/12)x=2

2A=5/12x=2

=>A=5/24x=1

=>x=1:5/24=24/5

 

5 tháng 4 2015

=>1/2.(5/12).x=1

5/24.x=1

x=1:5/24

x=24/5

lưu ý, 1/2.5/12 là tính xong phần 1/2.4 +...+1/10.12 rùi nhé

29 tháng 6 2015

1/

\(1+\frac{2014}{2}+...+\frac{4024}{2012}=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{2012}{2012}\right)\)

\(=2012+2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)

Phương trình đã cho  tương đương:

 \(\left(1+\frac{1}{2}+...+\frac{1}{2012}\right).503x=2012\left(1+\frac{1}{2}+...+\frac{1}{2012}\right)\)

\(\Leftrightarrow503x=2012\)

\(\Leftrightarrow x=4\)

2/ 

\(\frac{8}{1.9}+\frac{8}{9.17}+...+\frac{8}{49.57}+\frac{58}{57}+2x-2=2x+\frac{7}{3}+5x-\frac{8}{4}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{9}+\frac{1}{9}-\frac{1}{17}+...+\frac{1}{49}-\frac{1}{57}+\left(1+\frac{1}{57}\right)-2-\frac{7}{3}+\frac{8}{4}=5x\)

\(\Leftrightarrow\)\(5x=\frac{17}{3}\Leftrightarrow x=\frac{17}{15}\)

3/

Ta có: \(1+\frac{1}{n\left(n+2\right)}=\frac{n\left(n+2\right)+1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)

\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).....\left(1+\frac{1}{n\left(n+2\right)}\right)\)\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}.......\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)

\(=2.\frac{n+1}{n+2}