Question

cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\) với \(a,b,c\ne0,b\ne c\) chứng minh rằng \(\frac{a}{b}=\frac{a-c}{c-b}\)

Answer 1

Ta có: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\Rightarrow\frac{1}{c}\div\frac{1}{2}=\frac{1}{a}+\frac{1}{b}\)

\(\Rightarrow\frac{2}{c}=\frac{b+a}{ab}\)

\(\Rightarrow2ab=c\left(b+a\right)\)

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

\(\Rightarrow ab-bc=ac-ab\)

\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)(đpcm)