Đặt \(A=\) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{a\left(a+1\right)}=\frac{49}{100}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{a}-\frac{1}{a+1}=\frac{49}{100}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{a+1}=\frac{49}{100}\)
\(\)\(\Rightarrow\frac{1}{a+1}=\frac{1}{2}-\frac{49}{100}\)
\(\)\(\Rightarrow\frac{1}{a+1}=\frac{1}{100}\Rightarrow a+1=100\Rightarrow a=100-1\)
\(\Rightarrow a=99\)
Vậy \(a=99\)k cho mik nha :))