Question

Cho a,b,c,d \(\ne\) 0 và \(b^2=ac;c^2=bd\). Chứng minh \(\frac{a^2+b^2+c^2}{b^2+c^2+d^2}=\frac{\left(a+b+c\right)^2}{\left(b+c+d\right)^2}=\frac{a}{d}\)

Answer 1

Đề bài sai nhé

Đẳng thức này mới đúng: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}=\frac{a}{d}\)

Answer 2

\(\left\{{}\begin{matrix}b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\\c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\end{matrix}\right.\) \(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)

\(\Rightarrow\frac{a}{d}=\frac{abc}{bcd}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)