\(\sqrt{1+n^2+\frac{n^2}{\left(n+1\right)^2}}=\sqrt{\left(1+n-\frac{n}{n+1}\right)^2}=1+n-\frac{n}{n+1}\text{ }\left(n>0\right)\)

\(P==1+2015-\frac{2015}{2016}+\frac{2015}{2016}=2016\)

\(\left(1+n-\frac{n}{n+1}\right)^2=1+n^2+\frac{n^2}{\left(n+1\right)^2}+2\left(n-\frac{n}{n+1}-\frac{n^2}{n+1}\right)\)

\(=1+n^2+\frac{n^2}{\left(n+1\right)^2}+2.\frac{n^2+n-n-n^2}{n+1}\)

\(=1+n^2+\frac{n^2}{\left(n+1\right)^2}\)

Sao mr. Lazy làm nhiều thế?

Ta có:

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thế vô bài toán được

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

\(=1-\frac{1}{\sqrt{2016}}\)

Đặt 2015 = a Ta có :

\(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

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

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

\(=\sqrt{\frac{\left(a+1\right)^2+a^4+2a^3+2a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

\(=\sqrt{\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

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

= \(\frac{a^2+a+1}{a+1}+\frac{a}{a+1}=\frac{a^2+2a+1}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=2015+1=2016\)

Đề viết sai nha bạn phải là \(-\frac{2015^2}{2016^2}\)

\(=\sqrt{1+2015^2-\frac{2015^2}{2016^2}}+\frac{2015}{2016}\)

\(=\sqrt{\left(1+2015-\frac{2015}{2016}\right)^2}+\frac{2015}{2016}\)

\(=1+2015-\frac{2015}{2016}+\frac{2015}{2016}\)

\(=2016\)

tick cho mình nha

\(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\) 

                                                 =\(\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}\)

                                                    =\(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)

áp dụng vào biểu thức ta có\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)

                       =\(1-\frac{1}{\sqrt{2016}}\)

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