Question

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

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

Answer 1

Ta có :

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

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

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

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

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

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

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

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

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

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

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

Vậy \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)