Question

Rút gọn T = \(\sqrt{1+\dfrac{1}{2014^2}+\dfrac{1}{2015^2}}\)

Answer 1

\(\left\{{}\begin{matrix}x=2014\\T=\sqrt{1+\dfrac{1}{x^2}+\dfrac{1}{\left(x+1\right)^2}}\end{matrix}\right.\)

\(T=\sqrt{\dfrac{x^2+2x+1}{x^2}-\dfrac{2}{x}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\left(\dfrac{x+1}{x}\right)^2-2.\left(\dfrac{x+1}{x}\right)\left(\dfrac{1}{x}\right)+\dfrac{1}{\left(x+1\right)^2}}\)\(T=\sqrt{\left(\dfrac{x+1}{x}-\dfrac{1}{x+1}\right)^2}=\left|\dfrac{x+1}{x}-\dfrac{1}{x+1}\right|\)

\(T=\left|\dfrac{2015}{2014}-\dfrac{1}{2015}\right|=\dfrac{2015}{2014}-\dfrac{1}{2015}\)