Ta có:
\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\Leftrightarrow\dfrac{1}{c}.2=\dfrac{1}{a}+\dfrac{1}{b}\)
\(\Leftrightarrow\dfrac{2}{c}=\dfrac{a+b}{ab}\Leftrightarrow2ab=\left(a+b\right)c\)
\(\Leftrightarrow ab+ab=ac+bc\Leftrightarrow ab-bc=ac-ab\)
\(\Leftrightarrow b\left(a-c\right)=a\left(c-b\right)\Leftrightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\)