\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\ \Rightarrow\dfrac{1}{c}=\dfrac{a+b}{2ab}\\ \Rightarrow ac+bc=2ab\)
Giả sử \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\Rightarrow ac-ab=ab-bc\Rightarrow ac+bc=2ab\left(\text{luôn đúng}\right)\)
Vậy \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)