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