Đáp án:
$ x=39$
Giải thích các bước giải:
$\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{x.(x+2)}=\dfrac{20}{41}\\
\Leftrightarrow \dfrac{1}{2}\left (\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x.(x+2)} \right )=\dfrac{20}{41}\\
\Leftrightarrow \dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{20}{41}.2\\
\Leftrightarrow 1-\dfrac{1}{x+2}=\dfrac{40}{41}\\
\Leftrightarrow \dfrac{1}{x+2}=\dfrac{41}{41}-\dfrac{40}{41}\\
\Leftrightarrow \dfrac{1}{x+2}=\dfrac{1}{41}\\
\Leftrightarrow x+2=41\\
\Leftrightarrow x=39$
