VẬY A>B  

Chúc bạn học tốt ( -_- )

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

Ta có : \(A=\dfrac{n}{n}+1+\dfrac{n+1}{n+2}\left(n\ne0,n\ne-2\right)\)

\(=1+1+\dfrac{n+1}{n+2}\)

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

\(=\dfrac{2n+4+n+1}{n+2}=\dfrac{3n+5}{n+2}\)

Và \(B=\dfrac{2n+1}{2n+3}\)

Đặt \(n=4\) ta được :

\(A=\dfrac{3.4+5}{4+2}=\dfrac{17}{6}\)

\(B=\dfrac{2.4+1}{2.4+3}=\dfrac{9}{11}\)

Vì \(\dfrac{17}{6}>\dfrac{9}{11}\) nên \(A>B\)

Lời giải:
\(A=\frac{n}{n+1}+\frac{n+1}{n+2}=\frac{n(n+2)+(n+1)^2}{(n+1)(n+2)}=\frac{2n^2+4n+2}{n^2+3n+2}>1\) do $2n^2+4n+2> n^2+3n+2$ với mọi $n\in\mathbb{N}^*$

$B=\frac{2n+1}{2n+3}< 1$ do $2n+1< 2n+3$

Do đó $A>B$

= nhau nha ko tin thì hỏi cô giáo nha

n/2n + 3 < n + 2 / 2n + 1 

ta có 

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

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

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

3n/3x 2n+3 =n/2n+1

n/2n+1=3n/3x2n+3=3n/6n+3<3n/6n+2

n/2n+1<3n+1/6n+2