Question

cho a,b,c > 0 biet b²=ac ,c²=bd , d²=ce ,e²=df. chung minh \(\frac{a^5+b^5+c^5+d^5+e^5}{b^5+c^5+d^5+e^5+f^5}\)\(=\frac{a}{f}\)

 

Answer 1

Ta có: b² = ac => \(\frac{a}{b}=\frac{b}{c}\); c² = bd => \(\frac{b}{c}=\frac{c}{d}\);  d² = ce => \(\frac{c}{d}=\frac{d}{e}\); e² = df => \(\frac{d}{e}=\frac{e}{f}\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{e}{f}\)\(\Rightarrow\frac{a^5}{b^5}=\frac{b^5}{c^5}=\frac{c^5}{d^5}=\frac{d^5}{e^5}=\frac{e^5}{f^5}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{a^5}{b^5}=\frac{b^5}{c^5}=\frac{c^5}{d^5}=\frac{d^5}{e^5}=\frac{e^5}{f^5}=\frac{a^5+b^5+c^5+d^5+e^5}{b^5+c^5+d^5+e^5+f^5}\)(1)

Lại có: \(\frac{a^5}{b^5}=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}.\frac{d}{e}.\frac{e}{f}=\frac{a}{f}\)(2)

Từ (1), (2) \(\Rightarrow\frac{a^5+b^5+c^5+d^5+e^5}{b^5+c^5+d^5+e^5+f^5}=\frac{a}{f}\)(đpcm)