Question

So sánh:   \(\frac{n}{n+1}\)và \(\frac{n+2}{n+3}\)(n\(\in\)N)

Answer 1

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

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

vì \(\frac{-1}{n+1}

Answer 2

quy đồng lên thôi rồi so bình thường thôi sẽ có kết quả là <

Answer 3

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 n+1<n+3

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

=>\(1-\frac{1}{n+1}

Answer 4

Ta có : n/n+1=n.(n+3)/(n+1).(n+3)=n^2+3n/(n+1).(n+3)

n+2/n+3=(n+2).(n+1)/(n+1).(n+3)=(n^2+n)+(2n+2)/(n+1).(n+3)

Vì (n^2+n)+(2n+2)>n^2+3n nên (n^2+n)+(2n+2)/(n+1).(n+3)>n^2+3n/(n+1).(n+3)

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

Vậy n+2/n+3>n/n+1