Question

Cho \(a,b,c\in R\) và \(a,b,c\ne0\) thỏa mãn \(b^2=ac\) . Chứng minh rằng :

\(\dfrac{a}{c}=\dfrac{\left(a+2007b\right)^2}{\left(b+2007c\right)^2}\)

Answer 1

\(b^2=ac\Rightarrow\dfrac{b}{a}=\dfrac{c}{b}\)

Đặt :\(\dfrac{b}{a}=\dfrac{c}{b}=k\Rightarrow b=ak\)

\(c=bk\)

\(\Rightarrow c=akk=ak^2\)

VT\(=\dfrac{a}{c}=\dfrac{a}{ak^2}=\dfrac{1}{k^2}\)

VP \(=\dfrac{\left(a+2007b\right)^2}{\left(b+2007c\right)^2}=\dfrac{\left(a+2007ak\right)^2}{\left(b+2007bk\right)^2}\)

\(=\dfrac{\left[a\left(1+2007k\right)\right]^2}{\left[b\left(1+2007k\right)\right]^2}=\dfrac{a^2\left(1+2007k\right)^2}{b^2\left(1+2007\right)^2}=\dfrac{a^2}{b^2}=\dfrac{a^2}{\left(ak^2\right)}=\dfrac{a^2}{a^2k^2}=\dfrac{1}{k^2}\)

\(\Rightarrow VT=VP\Rightarrow\dfrac{a}{b}=\dfrac{\left(a+2007b\right)^2}{\left(b+2007c\right)^2}\)