n/n+2 > n/n+9 > n-2/n+9

=> n/n+2 > n-2/n+9

K mk nha,mk âm điểm rùi!huhu

n-2<n+9\(\Rightarrow\)\(\frac{n-2}{n+9}<\frac{n}{n+11}<\frac{n}{n+2}\)

Vậy \(\frac{n}{n+2}>\frac{n-2}{n+9}\)

Ta số phân số chung gian là \(\frac{n+1}{n+3}\)

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

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

Ủng hộ nhé ! 

 Ta có: \(\frac{n-2}{n+9}=\frac{n}{n+9}-\frac{2}{n+9}\)(n thuộc N*). Vì \(\frac{n}{n+8}>\frac{n}{n+9}\)nên  \(\frac{n}{n+8}>\frac{n}{n+9}>\frac{n}{n+9}-\frac{2}{n+9}\)

a) \(\frac{5}{9}=\frac{20}{36};\frac{1}{4}=\frac{9}{36}\)

\(\frac{20}{36}>\frac{9}{36}\Rightarrow\frac{5}{9}>\frac{1}{4}\)

\(\frac{72}{73}=\frac{4248}{4307};\frac{58}{59}=\frac{4234}{4307}\)

\(\frac{4248}{4307}>\frac{4234}{4307}\Rightarrow\frac{72}{73}>\frac{58}{59}\)

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

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

a,>

b,>

c,<

Ta có :

\(\frac{n-2}{n+9}=\frac{n}{2+9}-\frac{2}{2+9}\)\(\left(n\in N\text{*}\right)\)

Vì \(\frac{n}{n+8}>\frac{n}{n+9}\)

\(\Rightarrow\frac{n}{n+8}>\frac{n}{n+9}>\frac{n}{n+9}-\frac{2}{n+9}\)

\(\Leftrightarrow\frac{n}{n+8}>\frac{n}{n+9}>\frac{n-2}{n+9}\)

\(\frac{\Rightarrow n}{n+8}>\frac{n-2}{n+9}\)

\(\frac{n}{n+1}\)<\(\frac{n+2}{n+3}\) với n>=0 

h) Ta có: \(\frac{n+1}{n+2}=1-\frac{1}{n+2}\)

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

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

\(\Rightarrow1-\frac{1}{n+2}< 1-\frac{1}{n+4}\)\(\Rightarrow\frac{n+1}{n+2}< \frac{n+3}{n+4}\)

Ta có : \(\frac{n+1}{n+2}=1-\frac{1}{n+2}\)

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

Mà \(\frac{1}{n+2}>\frac{1}{n+4}\)

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