Question

[Mr Valentine]

So sánh:

A = n + 2016 / n

và B = n + 2017 / n +1 với n\(\in\)N

Answer 1

[Kaori Miyazono]

\(A=\frac{n+2016}{n}=\frac{n}{n}+\frac{2016}{n}\)

\(B=\frac{n+2017}{n+1}=\frac{n+1+2016}{n+1}=\frac{n+1}{n+1}+\frac{2016}{n+1}=1+\frac{2016}{n+1}\)

Vì \(1=1\);\(\frac{2016}{n}>\frac{2016}{n+1}\)nên \(1+\frac{2016}{n}>1+\frac{2016}{n+1}\)hay \(A>B\)

Vậy \(A>B\)

Answer 2

[Đinh Đức Hùng]

\(A=\frac{n+2016}{n}=\frac{n}{n}+\frac{2016}{n}=1+\frac{2016}{n}\)

\(B=\frac{n+2017}{n+1}=\frac{\left(n+1\right)+2016}{n+1}=1+\frac{2016}{n+1}\)

Vì \(n< n+1\) nên \(\frac{2016}{n}>\frac{2016}{n+1}\) => \(1+\frac{2016}{n}>1+\frac{2016}{n+1}\)

Do đó A > B