Question

Rút gọn:

P=\(\sqrt{1^2+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1^2+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1^2+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}\)

Mọi người giúp em nhé!!!!!

Answer 1

\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{a}\right)^2-\dfrac{2}{a}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(\dfrac{a+1}{a}\right)^2-2.\dfrac{a+1}{a}.\dfrac{a}{a+1}+\dfrac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\dfrac{1}{a}-\dfrac{1}{a+1}\right)^2}=1+\dfrac{1}{a}-\dfrac{1}{a+1}\left(a>0\right)\)

Áp dụng điều này vào bài toán , ta có :

\(P=\sqrt{1^2+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1^2+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+...+\sqrt{1^2+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}=2017+\dfrac{1}{2}-\dfrac{1}{2019}\)