Question

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

Answer 1

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

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

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

\(\Rightarrow2ab=ac+bc\)

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

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

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

Answer 2

Võ Nguyễn Thương Thương 

Answer 3

oOo Thiếu gia ác ma đừng hôn tôi oOo có gì ko