Question

Cho \(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) (với a,b,c khác 0, b khác c) chứng minh rằng \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)

Answer 1

\(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\Leftrightarrow\dfrac{1}{c}.2=\dfrac{1}{a}+\dfrac{1}{b}\)

\(\Leftrightarrow\dfrac{2}{c}=\dfrac{a+b}{ab}\Leftrightarrow2ab=\left(a+b\right)c\)

\(\Leftrightarrow ab+ab=ac+bc\)

\(\Leftrightarrow ab-bc=ac-ab\Leftrightarrow b\left(a-c\right)=a\left(c-b\right)\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\)

Bài này mình cũng đã trả lời rồi đấy ạ =))