Đặt\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
\(\Rightarrow k^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\)(1)
Lại có: \(k=\) \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\) \(\Rightarrow k^3=\left(\frac{a+b+c}{b+c+d}\right)^3\left(2\right)\)
Từ (1),(2)\(\Rightarrow\)\(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
Đúng 0
Bình luận (0)