Question

cho \(\frac{1}{c}\)=\(\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)(với a, b, c \(\ne\)0; b\(\ne\)0)

chứng minh rằng: \(\frac{a}{b}=\frac{a-c}{c-b}\)

Answer 1

Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)

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

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

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

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

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

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