Đặt : \(A=\dfrac{n+1}{n+5}\) và \(B=\dfrac{n+3}{n+4}\).

Ta có : \(A=\dfrac{n+1}{n+5}=\dfrac{n+5-4}{n+5}=\dfrac{n+5}{n+5}-\dfrac{4}{n+5}=1-\dfrac{4}{n+5}\)

Và : \(B=\dfrac{n+3}{n+4}=\dfrac{n+4-1}{n+4}=\dfrac{n+4}{n+4}-\dfrac{1}{n+4}=1-\dfrac{1}{n+4}\)

Cả \(A\) và \(B\) đều có hạng tử \(1\) nên ta so sánh : \(\dfrac{4}{n+5}\) và \(\dfrac{1}{n+4}\).

Quy đồng ta được : 

\(\dfrac{4\left(n+4\right)}{\left(n+5\right)\left(n+4\right)}=\dfrac{4n+16}{\left(n+5\right)\left(n+4\right)}\) và \(\dfrac{n+5}{\left(n+4\right)\left(n+5\right)}\).

Do mẫu bằng nhau nên ta so sánh tử, ta thấy : 

\(4n+16-\left(n+5\right)=4n+16-n-5=3n+11\).

Do \(n\) là số tự nhiên nên \(3n\ge0\), suy ra \(3n+11\ge11\).

Suy ra được : \(4n+16-\left(n+5\right)=3n+11\ge11>0\) nên \(4n+16>n+5\).

Do đó, \(\dfrac{4}{n+5}>\dfrac{4}{n+4}\Rightarrow1-\dfrac{4}{n+5}< 1-\dfrac{4}{n+4}\).

Vậy : \(A< B\) hay \(\dfrac{n+1}{n+5}< \dfrac{n+3}{n+4}\).

Ta có : \(\frac{n+1}{n+5}< \frac{n+1}{n+3}\)   mà \(\frac{n+1}{n+3}< \frac{n+2}{n+3}\)

\(\Rightarrow\)\(\frac{n+1}{n+5}< \frac{n+2}{n+3}\)\(,\forall\)\(n\in N\)

a) Theo đầu bài ta có:
\(\orbr{\begin{cases}\frac{n}{n+1}=\frac{n\left(n+4\right)}{\left(n+1\right)\left(n+4\right)}=\frac{n^2+2n+2n}{\left(n+1\right)\left(n+4\right)}\\\frac{n+1}{n+4}=\frac{\left(n+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+4\right)}=\frac{n^2+2n+1}{\left(n+1\right)\left(n+4\right)}\end{cases}}\)
Nếu \(n=0\Rightarrow2n=0< 1\Rightarrow\frac{n^2+2n+2n}{\left(n+1\right)\left(n+4\right)}< \frac{n^2+2n+1}{\left(n+1\right)\left(n+4\right)}\Rightarrow\frac{n}{n+1}< \frac{n+1}{n+4}\)
Nếu \(n\ge1\Rightarrow2n\ge2>1\Rightarrow\frac{n^2+2n+2n}{\left(n+1\right)\left(n+4\right)}>\frac{n^2+2n+1}{\left(n+1\right)\left(n+4\right)}\Rightarrow\frac{n}{n+1}>\frac{n+1}{n+4}\)

Ta có : \(\frac{n}{n+6}\)=\(1-\frac{6}{n+6}\)

           \(\frac{n+1}{n+7}\)=\(1-\frac{6}{n+7}\)

Vì \(\frac{6}{n+6}>\frac{6}{n+7}\)=> \(\frac{n}{n+6}< \frac{n+1}{n+7}\)Vì phần cần thêm vào càng lớn thì phần có sẵn càng nhỏ 

ủng hộ mik nhaaa

Ta có:

\(1-\frac{n}{n+6}=\frac{n+6}{n+6}-\frac{n}{n+6}=\frac{6}{n+6}.\)

\(1-\frac{n+1}{n+7}=\frac{n+7}{n+7}-\frac{n+1}{n+7}=\frac{6}{n+7}.\)

Vì \(n+6< n+7\)nên \(\frac{6}{n+6}>\frac{6}{n+7}\Leftrightarrow1-\frac{6}{n+6}< 1-\frac{6}{n+7}\Leftrightarrow\frac{n}{n+6}< \frac{n+1}{n+7}\)

k với!!!!!!!!!!!!

ta có: \(\frac{n}{n+3}=\frac{n\left(n+2\right)}{\left(n+3\right)\left(n+2\right)}=\frac{n^2+2n}{\left(n+3\right)\left(n+2\right)}\)

\(\frac{n+1}{n+2}=\frac{\left(n+1\right)\left(n+3\right)}{\left(n+2\right)\left(n+3\right)}=\frac{n^2+3n+n+3}{\left(n+2\right)\left(n+3\right)}\)

thấy rõ \(\frac{n^2+2n}{\left(n+3\right)\left(n+2\right)}<\frac{n^2+3n+n+3}{\left(n+3\right)\left(n+2\right)}\Rightarrow\frac{n}{n+3}<\frac{n+1}{n+2}\)

Ngoài ra bạn có thể sử dụng phương pháp so sánh phần bù

n+1/n+2<1
Suy ra n+1/n+2<n+2/n+1+2=n+2/n+3

\(\dfrac{n+1}{2n+3}\) < \(\dfrac{n+1}{2n+2}\) < \(\dfrac{n+2}{2n+2}\)

phân số n+1/n+2 lớn hơn