Question

[dophuonganh]

so sanh n/n+1va n+1/n+2

Answer 1

[DanAlex]

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

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

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

Answer 2

[Dũng Lê Trí]

\(\frac{n}{n+1};\frac{n+1}{n+2}\)

Ta có :

\(\frac{n}{n+1}=\frac{n\cdot\left(n+2\right)}{\left(n+1\right).\left(n+2\right)}\)

\(\frac{n+1}{n+2}=\frac{\left(n+1\right)\cdot\left(n+1\right)}{\left(n+1\right)\cdot\left(n+2\right)}\)

Bây giờ 2 phân số đều chung mẫu (n+1)*(n+2)

so sánh n*(n+2) và (n+1)*(n+1)

\(n.\left(n+2\right)=n^2+2n\)

\(\left(n+1\right)\cdot\left(n+1\right)=n\cdot\left(n+1\right)+1\cdot\left(n+1\right)\)

\(n\cdot\left(n+1\right)+1\cdot\left(n+1\right)=n^2+2n+1\)

\(n^2+2n< n^2+2n+1\)

=> \(\frac{n}{n+1}< \frac{n+1}{n+2}\)