a). n/n+1  < n+2/n+3 

b). n/n+3 > n−1/n+4 

c). n/2n+1 < 3n+1/6n+3 

k mk nha

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

=>n/n+1<n+2/n+3

vậy........

b)\(\frac{n}{n+3}>\frac{n}{n+4}>\frac{n-1}{n+4}\Rightarrow\frac{n}{n+3}>\frac{n}{n+4}\)

vậy.....

c)\(\frac{n}{2n+1}=\frac{3n}{6n+3}< \frac{3n+1}{6n+3}\)

vậy.......

a) \(\frac{n}{n+1}=1-\frac{1}{n+1};\frac{n+2}{n+3}=1-\frac{1}{n+3}\)
Vì \(\frac{1}{n+1}>\frac{1}{n+3}\)=) \(1-\frac{1}{n+1}< 1-\frac{1}{n+3}\)
=) \(\frac{n}{n+1}< \frac{n+2}{n+3}\)
b) Áp dụng tính chất : Nếu \(\frac{a}{b}< 1\)=) \(\frac{a}{b}< \frac{a+m}{b+m}\)
 Ta có : \(\frac{n-1}{n+4}< 1\)=) \(\frac{n-1}{n+4}< \frac{n-1+1}{n+4+1}=\frac{n}{n+5}< \frac{n}{n+3}\)
=) \(\frac{n-1}{n+4}< \frac{n}{n+3}\)

Mình mới lớp 5 nên không biết làm bài này.

Xin lỗi nha! Chúc bạn may mắn......mình chính là Đào Minh Tiến!

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

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

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

So sánh : \(n\cdot\left(n+3\right)\)và \(\left(n+2\right)\cdot\left(n+3\right)\)

\(n\cdot\left(n+3\right)=n^2+3n\)

\(\left(n+2\right)\cdot\left(n+3\right)=n^2+5n+6\)

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

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

b) \(\frac{n}{2n+1}\)và \(\frac{3n+1}{6n+3}\)

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

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

So sánh : \(n\cdot\left(6n+3\right)\)và \(\left(3n+1\right)\cdot\left(2n+1\right)\)

\(n\cdot\left(6n+3\right)=6n^2+3n\)

\(\left(3n+1\right)\cdot\left(2n+1\right)=6n^2+5n+1\)

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

\(\Leftrightarrow\frac{n}{2n+1}< \frac{3n+1}{6n+3}\)

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

                        \(=\frac{3n}{6n+3}\)

Đến đây so sánh tử số.

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

Xét 2 mẫu của phân số: \(6n+3=6n+3\)

Xét 2 tử số của hai phân số: \(3n+1>3n\)

\(\Rightarrow\frac{3n}{6n+3}< \frac{3n+1}{6n+3}\)(phân số nào cùng mẫu, có tử lớn hơn thì lớn hơn)

a)A=n/n+1=n/n+0/1

   B=n+2/n+3=n/n  +  2/3

ta có:0<2/3

=>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)     <

b)     <

c)     >

a) quy đồng : \(\frac{n}{2n+1}=\frac{3n}{6n+3}\)

Vì 3n < 3n + 1 => \(\frac{3n}{6n+3}< \frac{3n+1}{6n+3}\)hay \(\frac{n}{2n+1}< \frac{3n+1}{6n+3}\)

b) Ta có :

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

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

Vì \(\frac{1}{n+1}>\frac{1}{n+3}\)nên \(1-\frac{1}{n+1}< 1-\frac{1}{n+3}\)hay \(\frac{n}{n+1}< \frac{n+2}{n+3}\)

c)  giả sử \(\frac{n}{n+3}< \frac{n-1}{n+4}\)

\(\Leftrightarrow\frac{n\left(n+4\right)}{\left(n+3\right)\left(n+4\right)}< \frac{\left(n-1\right)\left(n+3\right)}{\left(n+3\right)\left(n+4\right)}\)

\(\Rightarrow n^2+4n< n^2+2n-3\)

\(\Rightarrow2n< -3\)( vô lí )

Vậy \(\frac{n}{n+3}>\frac{n-1}{n+4}\)