Question

so sánh

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

Answer 1

Ta có : \(\frac{n}{n+1}=\frac{n\left(n+3\right)}{\left(n+1\right)\left(n+3\right)}=\frac{n^2+3n}{n^2+3n+n+3}=\frac{n^2+3n}{n^2+4n+3}\)

             \(\frac{n+2}{n+3}=\frac{\left(n+2\right)\left(n+1\right)}{\left(n+3\right)\left(n+1\right)}=\frac{n^2+n+2n+2}{n^2+n+3n+3}=\frac{n^2+3n+2}{n^2+4n+3}\)

Vì  n² + 3n < n² + 3n + 2 => \(\frac{n^2+3n}{n^2+4n+3}<\frac{n^2+3n+2}{n^2+4n+3}\) =>  \(\frac{n}{n+1}<\frac{n+2}{n+3}\)