ta thấy:

\(\frac{n}{n+3}< 1\Rightarrow\frac{n}{n+3}< \frac{n+1}{n+4}< \frac{n+1}{n+2}\)

\(\Rightarrow\frac{n}{n+3}< \frac{n+1}{n+2}\)

vậy ...

Đặt A = \(\frac{n+1}{n+2}\)

=> \(\frac{1}{A}=\frac{n+2}{n+1}\)

=> \(\frac{1}{A}-1=\frac{n+2-n-1}{n+1}=\frac{1}{n+1}\)

Đặt B = \(\frac{n+3}{n+4}\)

=> \(\frac{1}{B}=\frac{n+4}{n+3}\)

=> \(\frac{1}{B}-1=\frac{n+4-n-3}{n+3}=\frac{1}{n+3}\)

Vì \(\frac{1}{n+1}>\frac{1}{n+3}\Rightarrow\frac{1}{A}-1>\frac{1}{B}-1\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)

Vậy \(\frac{n+1}{n+2}< \frac{n+3}{n+4}\)

Đặt \(A=\frac{n+1}{n+2}\)

\(\Rightarrow\frac{1}{A}=\frac{n+2}{n+1}\)

\(\Rightarrow\frac{1}{A}-1=\frac{n+2-n+1}{n+1}=\frac{1}{n+1}\)

Đặt \(B=\frac{n+3}{n+4}\)

\(\Rightarrow\frac{1}{B}=\frac{n+4}{n+3}\)

\(\Rightarrow\frac{1}{B}-1=\frac{n+4-n-3}{n+3}=\frac{1}{n+3}\)

Vì \(\frac{1}{n+1}>\frac{1}{n+3}\Rightarrow\frac{1}{A}-1>\frac{1}{B}-1\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)

Vậy \(\frac{n+1}{n+2}< \frac{n+3}{n+4}\)

ta co :

                        n/n+3=n+3-3/n+3=1-3/n+3

                        n+1/n+2=n+2-1/n+2=1-1/n+2

vi 3/n+3>1/n+2 nen n/n+3<n+1/n+2

Ta có : \(\frac{n}{n+1}=\frac{n\left(n+3\right)}{\left(n+1\right)\left(n+3\right)}=\frac{n^2+3n}{n^2+3n+n+3}=\frac{n^2+3n}{n^2+4n+3}\)

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

Vì  n² + 3n < n² + 3n + 2 => \(\frac{n^2+3n}{n^2+4n+3}<\frac{n^2+3n+2}{n^2+4n+3}\) =>  \(\frac{n}{n+1}<\frac{n+2}{n+3}\)

ta có :\(\left(n+1\right).\left(n+3\right)=n^2+4n+3\)

         \(n\left(n+2\right)=n^2+2n\)

=>\(\frac{n+1}{n+2}>\frac{n}{n+2}\)(vì có tích chéo lớn hơn)

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

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

Từ (*) và (**) có: \(\frac{n+1}{n+2}>\frac{n}{n+3}\)

\(\frac{n+1}{n+2}\) và \(\frac{n}{n+3}\)

\(PSTG:\frac{n}{n+2}\)

\(\frac{n+1}{n+2}>\frac{n}{n+2}\)

\(\frac{n}{n+2}>\frac{n}{n+3}\)

\(\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)

Ta có :   \(\left(-n-2\right).\left(-n-2\right)\)

\(=\left(-n-2\right).-n-\left(-n-2\right).2\)

\(=\left(-n\right).\left(-n\right)-2.\left(-n\right)-\left[-n.2-2.2\right]\)

\(=n^2+2n+2n+4\)

\(=n^2+4n+4\)( 1 ) 

\(\left(n+1\right)\left(n+3\right)\)

\(=\left(n+1\right).n+\left(n+1\right).3\)

\(=n^2+n+3n+3\)( 2 ) 

Từ ( 1 ) ; ( 2 ) 

\(\Rightarrow\left(-n-2\right)\left(-n-2\right)>\left(n+1\right)\left(n+3\right)\)

\(\Rightarrow\frac{n+1}{-n-2}>\frac{-n-2}{n+3}\)

Chúc bạn học tốt !!!!