Ta có:
\(\dfrac{a}{b}=\dfrac{a.\left(b+2017\right)}{b.\left(b+2017\right)}=\dfrac{a.b+a.2017}{b.\left(b+2017\right)}\) (1)
\(\dfrac{a+2017}{b+2017}=\dfrac{b.\left(a+2017\right)}{b.\left(b+2017\right)}=\dfrac{a.b+b.2017}{b.\left(b+2017\right)}\) (2)
Từ (1) và (2) suy ra:
+) Nếu a >b thì \(\dfrac{a.b+a.2017}{b.\left(b+2017\right)}>\dfrac{b.a+b.2017}{b.\left(b+2017\right)}\Rightarrow\dfrac{a}{b}>\dfrac{a+2017}{b+2017}\)
+) Nếu a <b thì \(\dfrac{a.b+a.2017}{b.\left(b+2017\right)}< \dfrac{a.b+b.2017}{b.\left(b+2017\right)}\) \(\Rightarrow\dfrac{a}{b}< \dfrac{a+2017}{b+2017}\)
+) Nếu a =b thì \(\dfrac{a.b+a.2017}{b.\left(b+2017\right)}=\dfrac{b.a+b.2017}{b.\left(b+2017\right)}\Rightarrow\dfrac{a}{b}=\dfrac{a+2017}{b+2017}\)