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

cho tớ l i k e trước nhé rồi tớ sẽ trả lời

Ta có: \(\frac{n}{n+1}=\frac{n\times n+2}{n+1\times n+2}\)
            \(\frac{n+1}{n+2}=\frac{n+1\times n+1}{n+2\times n+1}=\frac{n\times2}{n\times3}\)
=> n + 1/ n + 2 > n/n+1

có Q=3n+1/6n+3

=3n/(6n+3)+1/(6n+3)

=3n/3.(2n+1)+1/6n+3

=n/3n+1+1/6n+3

Ta có :

A = n / 2n + 1 = 3n / 3 ( 2n + 1 )  = 3n / 6n + 3

Vì 3n / 6n + 3 < 3n + 1/ 6n + 3 => A < B

Vậy A < B

\(\frac{n}{2n+1}\)=\(\frac{3.n}{3.\left(2n+1\right)}\)=\(\frac{3n}{6n+3}\)

Vì 6n+3=6n+3;3n<3n+1 nên \(\frac{n}{2n+1}\)<\(\frac{3n+1}{6n+3}\)

\(A=\dfrac{n}{2n+1}=\dfrac{n\left(6n+3\right)}{\left(2n+1\right)\left(6n+3\right)}\dfrac{6n^2+3n}{\left(2n+1\right)\left(6n+3\right)}\)

\(B=\dfrac{3n+1}{6n+3}=\dfrac{\left(3n+1\right)\left(2n+1\right)}{\left(6n+3\right)\left(2n+1\right)}=\dfrac{6n^2+5n+1}{\left(6n+3\right)\left(2n+1\right)}\)

Lại có :

\(6n^2+3n< 6n^2+5n+1\)

\(\Leftrightarrow A< B\)

$A=\frac{n}{2n+1}=\frac{n\left(6n+3\right)}{\left(2n+1\right)\left(6n+3\right)}\frac{6n^2+3n}{\left(2n+1\right)\left(6n+3\right)}$

$B=\frac{3n+1}{6n+3}=\frac{\left(3n+1\right)\left(2n+1\right)}{\left(6n+3\right)\left(2n+1\right)}=\frac{6n^2+5n+1}{\left(6n+3\right)\left(2n+1\right)}$

Lại có :

$6n^2+3n<6n^2+5n+1$

$\iff A<B$

A = n/2n+1 = 3n / 6n+3 < 3n+1/6n+3 = B

=> A < B