Ta có: \(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{x.\left(x+2\right)}=\frac{8}{7}\)
\(\Leftrightarrow3\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{x.\left(x+2\right)}\right)=\frac{8}{7}\)
\(\Leftrightarrow3.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x\left(x+2\right)}\right).\frac{1}{2}=\frac{8}{7}\)
\(\Leftrightarrow\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{8}{7}\)
\(\Leftrightarrow\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{x+2}\right)=\frac{8}{7}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{8}{7}:\frac{3}{2}=\frac{8}{7}.\frac{2}{3}=\frac{16}{21}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{16}{21}=\frac{-3}{7}\)
\(\Leftrightarrow\frac{-3}{-3\left(x+2\right)}=\frac{-3}{7}\)
\(\Leftrightarrow-3\left(x+2\right)=7\)
\(\Leftrightarrow x+2=\frac{7}{-3}\)
\(\Leftrightarrow x=\frac{7}{-3}-2=\frac{-13}{3}\)
Vậy \(x=\frac{-13}{3}\)
