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

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

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

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

1212;

1212;

1212.

k cho mình nhé.

1212

tk nhe@@@@@@@@@@@!!

aitk minh minh tk lai

bye

Ta so sánh từng số hạng : 

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

\(\frac{\sqrt{3}-\sqrt{2}}{2+3}=\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(2+3\right)\left(\sqrt{3}+\sqrt{2}\right)}=\frac{1}{\left(2+3\right)\left(\sqrt{2}+\sqrt{3}\right)}< \frac{1}{2}\)

..........................................................................................................................................................................................

\(\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}=\frac{\left(\sqrt{2015}-\sqrt{2014}\right)\left(\sqrt{2015}+\sqrt{2014}\right)}{\left(2014+2015\right)\left(\sqrt{2015}+\sqrt{2014}\right)}=\frac{1}{\left(2014+2015\right)\left(\sqrt{2015}+\sqrt{2015}\right)}< \frac{1}{2}\)

Vì mỗi số hạng của M đều nhỏ hơn 1/2 nên M < 1/2

Bài này mình làm chưa đúng nhé :) Để lát mình làm cách khác.