Ta có :
\(\frac{6}{1.3.5}+\frac{6}{3.5.7}+...+\frac{6}{15.17.19}-x=1\)
\(\Leftrightarrow\)\(\frac{6}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{15.17.19}\right)=x+1\)
\(\Leftrightarrow\)\(\frac{6}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{15.17}-\frac{1}{17.19}\right)=x+1\)
\(\Leftrightarrow\)\(\frac{6}{4}\left(\frac{1}{3}-\frac{1}{323}\right)=x+1\)
\(\Leftrightarrow\)\(\frac{6}{4}.\frac{320}{969}=x+1\)
\(\Leftrightarrow\)\(\frac{160}{323}=x+1\)
\(\Leftrightarrow\)\(x=\frac{160}{323}-1\)
\(\Leftrightarrow\)\(x=\frac{-163}{323}\)
Vậy \(x=\frac{-163}{323}\)
Chúc bạn học tốt ~