Question

cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)vơi a,b,c \(\ne\) 0; b\(\ne\) c chứng minh rằng \(\frac{a}{b}=\frac{a-c}{c-b}\)

Answer 1

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

\(=\frac{1}{c}\times2=\frac{1}{a}+\frac{1}{b}\)

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

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

= \(c\left(b+a\right)=ab\times2\)

= cb +ca = ab+ab

= ab - cb = ac-ab

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

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

Answer 2

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

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

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

\(2ab=c\left(a+b\right)\)

\(ab+ab=ac+bc\)

\(ab-bc=ac-ab\)

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

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