Question

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

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

Answer 1

theo bài ra ta có:

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

=> \(\frac{1}{c}=\frac{1}{2}.\frac{b+a}{ab}\\ \Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\)

=> 2ab = c( a+b )

=> ab + ab = ca + cb

<=> ca+cb = ab+ab

=> ca - ab = ab - cb

=> a( c - b ) = b( a - c )

=> \(\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)