Ta có \(1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}=\left(a+1-\dfrac{1}{a+1}\right)^2\Rightarrow\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=1+a-\dfrac{1}{a+1}\)
Áp dụng cái này vào bài, ta có
A=\(1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=100-\dfrac{1}{100}=\dfrac{9999}{100}\)