\(D=\frac{2.2014}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2014}}\)

\(D=\frac{2.2014}{\frac{2}{2}+\frac{1}{\frac{2.3}{2}}+...+\frac{1}{\frac{2015.2014}{2}}}\)

\(D=\frac{2.2014}{\frac{2}{2}+\frac{2}{2.3}+...+\frac{2}{2014.2015}}\)

\(D=\frac{2015}{\frac{1}{2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}}\)

\(D=\frac{2014}{\frac{1}{2}+\frac{1}{2}-\frac{1}{2015}}\)

\(D=\frac{2.2014}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2014}}\)

\(D=\frac{2.2014}{\frac{1}{\frac{\left(1+1\right).1}{2}}+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2014+1\right).2014}{2}}}\)

\(D=\frac{2.2014}{\frac{2}{1.2}+\frac{2}{3.2}+\frac{2}{4.3}+\frac{2}{2015.2014}}\)

\(D=\frac{2.2014}{2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)}\)

\(D=\frac{2014}{\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)}\)

\(D=\frac{2014}{\left(1-\frac{1}{2015}\right)}\)

\(D=\frac{2014}{\frac{2014}{2015}}\)

\(D=\frac{2014.2015}{2014}\)

\(D=2015\)

Tham khảo nhé~

Nhân cả 2 với (\(\sqrt{2015^2-1}\)+\(\sqrt{2014^2-1}\))

A = 2015^2 -1 -2014^2 + 1 = (2014 + 1)^2 -2014^2 = 2.2014 + 1

B = 2.2014

=> A = B + 1

Ta có :

\(A=\sqrt{2013^2+2013^2.2014^2+2014^2}\)

\(=\sqrt{\left(2013.2014\right)^2+2013.\left(2014-1\right)+\left(2013+1\right).2014}\)

\(=\sqrt{\left(2013.2014\right)^2+2013.2014-2013+2014+2014.2013}\)

\(=\sqrt{\left(2013.2014\right)^2+2.2013.2014.1+1^2}\)

\(=\sqrt{\left(2013.2014+1\right)^2}\)

\(=2013.2014+1\in N\)

Vậy ...

Ta có: \(A=\sqrt{2013^2+2013^2.2014^2+2014^2}\)

<=>\(A=\sqrt{\left(2014^2+2013^2-2.2013.3014\right)+2.2013.2014+\left(2013.2014\right)^2}\)

<=>\(A=\sqrt{\left(2014-2013\right)^2+2.2013.2014+\left(2013.2014\right)^2}\)

<=>\(A=\sqrt{1+2.2013.2014+\left(2013.2014\right)^2}\)

<=>\(A=\sqrt{\left(2013.2014+1\right)^2}\)

<=>A=2013.2014+1

<=>A=4054183

Vậy A là số tự nhiên

Frrrr cchhaaaa