Đặt \(\frac{a}{12}=\frac{b}{9}=\frac{c}{5}=k\left(k\in Z\right)\)
\(\Rightarrow a=12k;b=9k;c=5k\)
\(\Rightarrow a.b.c=540k^3=20\)
\(\Rightarrow k^3=\frac{1}{27}\Rightarrow k=\frac{1}{3}\)
\(\Rightarrow a=4;b=3;c=\frac{5}{3}\)
#)Giải :
Đặt \(\frac{a}{12}=\frac{b}{9}=\frac{c}{5}=k\Rightarrow\hept{\begin{cases}a=12k\\b=9k\\c=5k\end{cases}\Rightarrow a.b.c=12k.9k.5k=540k^3=20\Rightarrow k^3=\frac{1}{27}\Rightarrow k=\frac{1}{3}}\)
\(\Rightarrow\hept{\begin{cases}\frac{a}{12}=\frac{1}{3}\\\frac{b}{9}=\frac{1}{3}\\\frac{c}{5}=\frac{1}{3}\end{cases}\Rightarrow\hept{\begin{cases}a=4\\b=3\\c=\frac{5}{3}\end{cases}}}\)
Vậy ...
+ Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a}{12}=\frac{b}{9}=\frac{c}{5}=\frac{a+b+c}{12+9+5}=\frac{20}{26}=\frac{10}{13}\)
Suy ra \(\frac{a}{12}=\frac{10}{13}\Rightarrow a=\frac{120}{13}\)
\(\frac{b}{9}=\frac{10}{13}\Rightarrow b=\frac{90}{13}\)
\(\frac{c}{5}=\frac{10}{13}\Rightarrow c=\frac{50}{13}\)
Vậy \(a=\frac{120}{13}\); \(b=\frac{90}{13}\); \(c=\frac{50}{13}\)