=>b^3=abc
=>c^3=bcd
=>a^3+b^3+c^3/b^3+c^3+d^3=a^3+abc+bcd/d^3+abc+bcd
=>
ta có:
+ b2 = ac
\(\Rightarrow\)\(\frac{a}{b}\)=\(\frac{b}{c}\)
+ c2 = bd
\(\Rightarrow\)\(\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)=\(\frac{a.b.c}{b.c.d}\)=\(\frac{a}{d}\)(1)
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)=\(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)(2)
Từ (1) và (2) => \(\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)
k mik nha
+, \(b^2=ac\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}\left(1\right)\)
+, \(c^2=bd\)
\(\Rightarrow\frac{b}{c}=\frac{c}{d}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)\(=\frac{a.b.c}{b.c.d}=\frac{a}{d}\)\(\left(3\right)\)
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Leftrightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)\(\left(4\right)\)
Từ \(\left(3\right),\left(4\right)\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)\(\left(đpcm\right)\)