Ta có : \(\sqrt{1+\dfrac{1}{a}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{a}\right)^2-\dfrac{2}{a}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(\dfrac{1+a}{a}\right)^2-2.\dfrac{a+1}{a}.\dfrac{1}{a+1}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{a}-\dfrac{1}{a+1}\right)^2}=\text{ |}1+\dfrac{1}{a}-\dfrac{1}{a+1}\text{ |}\) Áp dụng điều này vào bài toán , ta có :
\(A=\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}}=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{n-1}-\dfrac{1}{n}=n-2+\dfrac{1}{2}-\dfrac{1}{n}=\dfrac{n^2-1}{n}-\dfrac{3}{2}\)
