áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{12a-15b}{7}=\frac{20c-12a}{9}=\frac{15b-20c}{11}=\frac{12a-15b+20c-12a+15b-20c}{7+9+11}=0\)
\(\frac{12a-15b}{7}=0\Rightarrow12a=15b\Rightarrow\frac{a}{15}=\frac{b}{12}\Rightarrow\frac{a}{5}=\frac{b}{4}\)(1)
\(\frac{20c-12a}{9}=0\Rightarrow20c=15a\Rightarrow\frac{a}{20}=\frac{c}{12}\Rightarrow\frac{a}{5}=\frac{c}{3}\)(2)
\(\frac{15b-20c}{11}=0\Rightarrow15b=20c\Rightarrow\frac{b}{20}=\frac{c}{15}\Rightarrow\frac{b}{4}=\frac{c}{3}\)(3)
từ (1),(2),(3) => \(\frac{a}{5}=\frac{b}{4}=\frac{c}{3}=\frac{a+b+c}{5+4+3}=\frac{48}{12}=4\)(t/c dãy tỉ số bằng nhau)
\(\frac{a}{5}=4\Rightarrow a=20,\frac{b}{4}=4\Rightarrow b=16,\frac{c}{3}=4\Rightarrow c=12\)
Vậy a=20, b=16, c=12
Áp dụng tc của dãy tỉ số bằng nhau :
\(\frac{12a-15b}{7}=\frac{20c-12a}{9}=\frac{15b-20c}{11}=\frac{12a-15b+20c-12a+15b-20c}{7+9+11}=\frac{0}{27}=0\)
\(=>\hept{\begin{cases}12a-15b=0=>12a=15b=>\frac{a}{5}=\frac{b}{4}\\20c-12a=0=>20c=12a=>\frac{c}{3}=\frac{a}{5}\\15b-20c=0=>15b=20c=>\frac{c}{3}=\frac{b}{4}\end{cases}=>\frac{a}{5}=\frac{b}{4}=\frac{c}{3}}\)
Đặt \(\frac{a}{5}=\frac{b}{4}=\frac{c}{3}=k=>\hept{\begin{cases}a=5k\\b=4k\\c=3k\end{cases}}\)
Thay vào : \(a+b+c=5k+4k+3k=12k=48=>k=4\)
\(=>\hept{\begin{cases}a=5k=5.4=20\\b=4k=4.4=16\\c=3k=3.4=12\end{cases}}\)
Vậy...
