Question

Cho \(\dfrac{1}{c}=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(a;b;c\ne0;b\ne c\right).\) Chứng minh: \(\dfrac{a}{b}=\dfrac{a-c}{c-b}\)

Answer 1

Ta có: \(\frac{1}{c}=\frac12\left(\frac{1}{a}+\frac{1}{b}\right)\)
=>\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}:\frac12=\frac{2}{c}\)

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

=>c(a+b)=2ab

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

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

\(=\frac{a^2-ab}{ab-b^2}=\frac{a\left(a-b\right)}{b\left(a-b\right)}=\frac{a}{b}\)