\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{x\left(x+2\right)}=\frac{16}{99}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{16}{99}\)
\(\Rightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x\left(x+2\right)}=\frac{32}{99}\)
=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
=> \(\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
=> \(\frac{1}{x+2}=\frac{1}{99}\)
=> x + 2 = 99
=> x = 97
Vậy x = 97 là giá trị cần tìm
\(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{x\times\left(x+2\right)}\)
\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{x\times\left(x+2\right)}\right)\)
\(=\frac{1}{2}\times\left(\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+\frac{9-7}{7\times9}+...+\frac{x+2-x}{x\times\left(x+2\right)}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{x+2}\right)\)
\(=\frac{1}{6}-\frac{1}{2\times\left(x+2\right)}=\frac{16}{99}\)
\(\Leftrightarrow\frac{1}{2\times\left(x+2\right)}=\frac{1}{6}-\frac{16}{99}=\frac{1}{198}\)
\(\Leftrightarrow2\times\left(x+2\right)=198\)
\(\Leftrightarrow x+2=99\)
\(\Leftrightarrow x=97\)
