giúp mik với mik cần gấp

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}

phân số n+1/n+2 lớn hơn

Thiếu đề hay xao ý bn

a,   <                b, >                 c, không biết

em mới hoc lớp 4 thôi

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