Question

cho \(\dfrac{1}{c}=\dfrac{1}{2}.\left(\dfrac{1}{a}+\dfrac{1}{c}\right)vớia,b,c\)khác 0 b khác 0

cm rằng\(\dfrac{a}{b}=\dfrac{a}{c}-\dfrac{c}{d}\)

Answer 1

Đề sai t sửa lại : Cho \(\dfrac{1}{c}=\dfrac{1}{2}.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\). CMR \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)

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

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

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

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

\(\Leftrightarrow2.ab=c\left(a+b\right)\)

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

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

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

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{a-c}{c-b}\left(đpcm\right)\)