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

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

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

Vậy \(\frac{n+1}{n+4}>\frac{n}{n+3}\)

a: so sánh với 1

64/85 < 73/81

b: so sánh với 1 

n + 1/n+2 > n/ n+3

c: so sánh với 1

64/65 > 60/61

d: so sánh với 1

99/97 < 88/86

n+1/n+2<1
Suy ra n+1/n+2<n+2/n+1+2=n+2/n+3

                         Giải

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

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

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

Vậy \(\frac{n+2}{n}>\frac{n+3}{n+1}\)

a: \(\dfrac{-8}{31}=\dfrac{-8\cdot101}{31\cdot101}=\dfrac{-808}{3131}\)

\(\dfrac{-789}{3131}=\dfrac{-789}{3131}\)

b: Thiếu phân số thứ hai rồi bạn

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

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

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

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

Mà \(\frac{1}{n+2}>\frac{1}{n+4}\)

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

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

\(1-\frac{n-1}{n+4}=\frac{n+4}{n+4}-\frac{n-1}{n+4}=\frac{n+5}{n+4}\)

Mà \(\frac{2}{n+2}1\)nên \(\frac{2}{n+2}

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

\(\frac{n-1}{n+4}=\frac{n+4-5}{n+4}=\frac{n+4}{n+4}-\frac{5}{n+4}=1-\frac{5}{n+4}\)

Ta có: \(\frac{2}{n+2}=\frac{\left(n+4\right)2}{\left(n+4\right)\left(n+2\right)}=\frac{2n+8}{n^2+2n+4n+8}\)

\(\frac{5}{n+4}=\frac{\left(n+2\right)5}{\left(n+2\right)\left(n+4\right)}=\frac{5n+10}{n^2+4n+2n+8}\)

Vì \(\frac{2n+8}{n^2+2n+4n+8}