K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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

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

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

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

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

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

=> \(2x=4\)

=> \(x=2\)

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 ( -_- )

10 tháng 2 2020

Đặt \(K=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+x}\)

\(=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{x\left(x+1\right)}{2}}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

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

Phương trình trở thành: \(2x:\left(2-\frac{2}{x+1}\right)=2020\)

\(\Leftrightarrow2x:\frac{2x}{x+1}=2020\Leftrightarrow x+1=2020\Leftrightarrow x=2019\)

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 ?

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