Question

cho b²=ac

cm\(\dfrac{a}{c}\)=\(\dfrac{\left(a+2019b\right)^2}{\left(b+2019c\right)^2}\)

Answer 1

Ta có:

\(b^2=ac\)

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

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\b=ck\end{matrix}\right.\)

Ta lại có:

\(\dfrac{\left(a+2019b\right)^2}{\left(b+2019c\right)^2}=\dfrac{\left(bk+2019b\right)^2}{\left(ck+2019c\right)^2}=\dfrac{\left(ck^2+2019ck\right)^2}{\left(ck+2019c\right)^2}=\left[\dfrac{k\left(ck+2019c\right)}{ck+2019c}\right]^2=k^2=\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}\)

Vậy ta có ĐPCM.