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

\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left(2x-2\right)\cdot2x}=\frac{1}{8}\)

\(\Rightarrow\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left(2x-2\right)\cdot2x}=\frac{2}{8}\)

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

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

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

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

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

\(\frac{1}{2.4}\)\(\)

\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right).2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{11}{24}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)

\(\Rightarrow2x=24\)

\(\Rightarrow x=12\)

\(\text{Sửa đề:}\)

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

\(\text{Đặt biểu thức là A:}\)

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

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}=\frac{1}{8}\times2=\frac{1}{4}\)

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

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

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

\(A=\frac{1}{2}\times\frac{1}{2}-\frac{1}{2}\times\frac{1}{2x}=\frac{1}{8}\)

\(A=\frac{1}{4}-\frac{1}{4x}=\frac{1}{8}\)

\(\Rightarrow\frac{1}{4x}=\frac{1}{4}-\frac{1}{8}=\frac{2}{8}-\frac{1}{8}=\frac{1}{8}\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=8\div4=2\)

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!

Thanks you so much.

Thùy Anh tính sai rùi, dòng 2 phải nhân cả tử và mẫu cùng zới 2 mò, ko nhận ra ak mng ưi

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

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

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

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

\(=2.\frac{x-1}{2x}\)

\(=\frac{x-1}{x}\)

đề thiếu

1/1.3 - 1/2.4 + 1/3.5 - 1/4.6 + ...+ 1/97.99 - 1/98.100 + 3lxl = 1 

tách ra trc đầu tiên tính phần : 1/1.3 - 1/2.4 + 1/3.5 - 1/4.6 + ...+ 1/97.99 - 1/98.100

tách số lẻ và số chẵn ra 

(1/1.3+1/3.5+...+1/57.97+1/97.99)-(1/2.4+1/4.6+...1/98.100)

tính từng vế  vế đầu kết qu3 vế lẻ là : 49/99 

kết quả vế chẵn là 49/200

thì bài đó sẻ thành : 49/99+49/200+3lxl=1 

còn lại tự tinh nha 

\(x-\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}\right)=\dfrac{1}{100}\)

\(x-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)=\dfrac{1}{100}\)

\(x-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=\dfrac{1}{100}\)

\(x=\dfrac{51}{200}\)

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

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

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

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

=> 1/2x = 1/4

=> 2x = 4

x = 4 : 2

x = 2

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

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

\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{8}.2=\frac{1}{4}\)

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

\(\Rightarrow2x=4\Leftrightarrow x=2\)

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

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

\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{8}:\frac{1}{2}\)

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

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

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)

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

i dont know