Question

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

c/m:\(\dfrac{a}{b}=\)\(\dfrac{a-c}{c-b}\)

Answer 1

giúp mik với

Answer 2

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

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

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

\(\Leftrightarrow2ab=c\left(a+b\right)\)

\(\Leftrightarrow ab+ab-ca-cb=0\)

\(\Leftrightarrow ab-ca=cb-ab\)

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

hay \(\dfrac{a}{b}=\dfrac{c-a}{b-c}\)

hay \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)(đpcm)

Answer 3