Ta có \(\frac{17}{3}=5+\frac{2}{3}=5+\frac{1}{\frac{3}{2}}=5+\frac{1}{1+\frac{1}{2}}\)

=> m=5;n=1;p=2

Mk sắp phải đi hc rồi, làm câu đầu thôi nha.

Bài 1:

Ta có: \(\left|x+\frac{5}{6}\right|-\frac{1}{2}=\frac{-3}{7}\)

\(\Rightarrow\left|x+\frac{5}{6}\right|=\frac{1}{14}\)

\(\Rightarrow x+\frac{5}{6}=\frac{1}{14}\) hoặc \(x+\frac{5}{6}=\frac{-1}{14}\)

Với \(x+\frac{5}{6}=\frac{1}{14}\Rightarrow x=\frac{-16}{21}\)

Với \(x+\frac{5}{6}=\frac{-1}{14}\Rightarrow x=\frac{-19}{21}\)

Vậy \(x=\frac{16}{21}\) hoặc \(x=\frac{-19}{21}\).

Ta có : \(\frac{17}{3}=5+\frac{2}{3}=5+\frac{1}{\frac{3}{2}}=5+\frac{1}{1+\frac{1}{2}}\)

Nên suy ra : m = 5 ; n = 1 ; p = 2

1

1

\(\frac{3}{5}< \frac{7}{10}< \frac{7}{9}< \frac{8}{9}< \frac{16}{17}< 1< \frac{18}{17}\)

1/ Trước hết ta chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)

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

Áp dụng : 

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

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

Với mọi \(n\ge2\)

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

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

Lại có : \(\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}< \frac{2}{\sqrt{n}+\sqrt{n-1}}=\frac{2\left(\sqrt{n}-\sqrt{n-1}\right)}{\left(\sqrt{n}+\sqrt{n-1}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\)

                                     \(=2\left(\sqrt{n}-\sqrt{n-1}\right)\) (2)

Từ (1) và (2) suy ra \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \frac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)

Áp dụng với n = 2,3,4,...,100 được đpcm.

\(5+\frac{1}{1+\frac{1}{2}}=\frac{17}{3}\)

n=1

\(\frac{17}{3}=5+\frac{2}{3}=5+\frac{1}{\frac{3}{2}}=5+\frac{1}{1+\frac{1}{2}}\\ \) vậy n=1

Ta có: \(M=\frac{2014^2+1^2}{2014.1}+\frac{2013^2+2^2}{2013.2}+\frac{2012^2+3^2}{2012.3}+...+\frac{1008^2+1007^2}{1008.1007}\)

\(=\frac{2014}{1}+\frac{1}{2014}+\frac{2013}{2}+\frac{2}{2013}+\frac{2012}{3}+\frac{3}{2013}+...+\frac{1008}{1007}+\frac{1007}{1008}\)

\(=\frac{2014}{1}+\frac{2013}{2}+...+\frac{1}{2014}\)

\(=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)\)

\(=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+\frac{2015}{2015}\)

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

\(\Rightarrow\frac{M}{N}=\frac{2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}}=2015\)