Question

[Tran Le Hoang Vu]

Cho a,bϵN*. Hay so sanh

\(\dfrac{a+n}{b+n}\) va \(\dfrac{a}{b}\)

Answer 1

[HT.Phong (9A5)]

TH1: với \(a>b\) 

\(\dfrac{a}{b}-1=\dfrac{a-b}{b}\)

\(\dfrac{a+n}{b+n}-1=\dfrac{a+n-\left(b+n\right)}{b+n}=\dfrac{a-b}{b+n}\)

Mà: \(b+n>b\)

\(\Rightarrow\dfrac{a-b}{b+n}< \dfrac{a-b}{b}\)

\(\Rightarrow\dfrac{a+n}{b+n}-1< \dfrac{a}{b}-1\)

\(\Rightarrow\dfrac{a+n}{b+n}< \dfrac{a}{b}\)

TH2: với \(a< b\)

\(1-\dfrac{a}{b}=\dfrac{b-a}{b}\)

\(1-\dfrac{a+n}{b+n}=\dfrac{\left(b+n\right)-\left(a+n\right)}{b+n}=\dfrac{b-a}{b+n}\)

Mà: \(b+n>b\)

\(\Rightarrow\dfrac{b-a}{b+n}< \dfrac{b-a}{b}\)

\(\Rightarrow1-\dfrac{a+n}{b+n}< 1-\dfrac{a}{b}\)

\(\Rightarrow\dfrac{a+n}{b+n}>\dfrac{a}{b}\) 

TH3: \(a=b\)

\(\dfrac{a+n}{b+n}=\dfrac{a+n}{a+n}=1\)

\(\dfrac{a}{b}=\dfrac{a}{a}=1\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{a+n}{b+n}=1\)