Question

[Phan Hải Đăng]

Cho a, b, c, d, thỏa mãn \(b^2=ac\)\(c^2=bd\)và \(b^3-2018c^3-2019d^3\ne0\)

CM\(\frac{a^3-2018b^3-2019c^3}{b^3-2018c^3-2019d^3}=\frac{a}{d}\)R 

Answer 1

[ST]

\(\hept{\begin{cases}b^2=ac\\c^2=bd\end{cases}\Rightarrow\hept{\begin{cases}\frac{a}{b}=\frac{b}{c}\\\frac{b}{c}=\frac{c}{d}\end{cases}\Rightarrow}\frac{a}{b}=\frac{b}{c}=\frac{c}{d}}\)

=>\(\frac{a^3}{b^3}=\frac{2018b^3}{2018c^3}=\frac{2019c^3}{2019d^3}=\frac{a^3-2018b^3-2019c^3}{b^3-2018c^3-2019d^3}\left(1\right)\)

Mà \(\frac{a^3}{b^3}=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\left(2\right)\)

Từ (1) và (2) => đpcm