Ta có : \(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{a}\right)^2-\dfrac{2}{n}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(\dfrac{a+1}{a}\right)^2-2.\dfrac{a+1}{a}.\dfrac{1}{a+1}+\left(\dfrac{1}{a+1}\right)^2}=\sqrt{\left(1+\dfrac{1}{a}-\dfrac{1}{a+1}\right)^2}=1+\dfrac{1}{a}-\dfrac{1}{a+1}\left(a>0\right)\) Áp dụng điều này vào bài toán trên , ta được :
\(P=\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}\) \(P=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}\)
\(P=98+\dfrac{1}{2}-\dfrac{1}{100}\)
\(P=\dfrac{9849}{100}\)
Với a+b+c =0 (a,b,c \(\ne\) 0) , ta có:
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\)
Áp dụng cho từng thừa số của P, ta có:
\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}=\sqrt{\left(1+\dfrac{1}{2}+\dfrac{1}{-3}\right)-2\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)}\)
\(=\sqrt{\left(1+\dfrac{1}{2}-\dfrac{1}{3}\right)^2-2.\dfrac{3-2-1}{6}}=\left|1+\dfrac{1}{2}-\dfrac{1}{3}\right|=1+\dfrac{1}{2}-\dfrac{1}{3}\)
Tương tự :\(\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}=1+\dfrac{1}{3}-\dfrac{1}{4}\)
\(\sqrt{1+\dfrac{1}{99^2}+\dfrac{1}{100^2}}=1+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow P=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(=1+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{149}{100}\)
