Đáp án:
$x=2$
Giải thích các bước giải:
$\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{(2x-2).2x}=\dfrac{1}{8}\\
\Leftrightarrow 2.\left ( \dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{(2x-2).2x} \right )=\dfrac{1}{8}.2\\
\Leftrightarrow \dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{(2x-2).2x}=\dfrac{1}{4}\\
\Leftrightarrow \dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\\
\Leftrightarrow \dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\\
\Leftrightarrow 1-\dfrac{1}{x}=\dfrac{1}{2}\\
\Leftrightarrow \dfrac{1}{x}=1-\dfrac{1}{2}=\dfrac{1}{2}\\
\Leftrightarrow x=2$
