Giải:
Ta có:
\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(\Leftrightarrow\dfrac{1}{c}\div\dfrac{1}{2}=\dfrac{1}{a}+\dfrac{1}{b}\)
\(\Leftrightarrow\dfrac{1}{c}.\dfrac{2}{1}=\dfrac{a+b}{ab}\)
\(\Leftrightarrow\dfrac{2}{c}=\dfrac{a+b}{ab}\)
\(\Leftrightarrow2ab=ac+bc\left(1\right)\)
Lại có:
\(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)
\(\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Leftrightarrow ac-ab=ab-bc\)
\(\Leftrightarrow2ab=ac+bc\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Leftrightarrow\) Nếu \(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\Leftrightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\) (Đpcm)
