\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{8}{9}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{8}{9}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{9}\)
=>x+1=9
hay x=8
\(\Leftrightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{8}{9}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{8}{9}\)
\(\Leftrightarrow-\dfrac{1}{x+1}=\dfrac{8}{9}-1\)
\(\Leftrightarrow-\dfrac{1}{x+1}=-\dfrac{1}{9}\)
\(\Leftrightarrow x+1=9\)
\(\Leftrightarrow x=8\)