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

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

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

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

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

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

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

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

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

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

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

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{4}\)

\(\Leftrightarrow2x=4\\ \Leftrightarrow x=2\left(tm\right)\)

\(\Leftrightarrow\dfrac{1}{4}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.......+\dfrac{1}{\left(x-1\right)x}\right)=\dfrac{1}{8}\)   ( 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)

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

2 . 1/8 = 2/2.4 + 2/4.6 + ...+ 2/((2x -2).2x)

1/4 = 4-2/2.4 + 6-4/4.6 + ... + 2x-(2x-2)/(2x-2)+2x

1/4 = 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2x -2 - 1/2x

1/4 = 1/2 - 1/2x

1/4 = 2x-2/2.2x

Tự làm tiếp nhé

Mong Bạn Bấm Cho mik

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

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

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

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

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

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

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

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

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

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

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

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

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

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

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

\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{1}{4}\)

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

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-1}{2x}=\dfrac{-1}{4}\)

\(\Rightarrow x=2\)

Ta có: \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)

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

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

\(\Rightarrow8\left(x-1\right)=4x\Rightarrow8x-8=4x\Rightarrow4x=8\Rightarrow x=2\)

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

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

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

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

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

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

\(\Rightarrow2x=24\)

\(\Rightarrow x=12\)

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

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

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

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

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

\(A=\dfrac{1}{2}.\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{2x}=\dfrac{1}{8}\)

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

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

\(\Rightarrow4x=8\)

\(\Rightarrow x=8:4=2\)

ai giup ming vs