Giải:
Ta có: \(\dfrac{6}{7}a=\dfrac{9}{11}b=\dfrac{2}{3}c\Rightarrow\dfrac{a}{\dfrac{7}{6}}=\dfrac{b}{\dfrac{11}{9}}=\dfrac{c}{\dfrac{3}{2}}\)
Đặt \(\dfrac{a}{\dfrac{7}{6}}=\dfrac{b}{\dfrac{11}{9}}=\dfrac{c}{\dfrac{3}{2}}=k\Rightarrow\left\{{}\begin{matrix}a=\dfrac{7}{6}k\\b=\dfrac{11}{9}k\\c=\dfrac{3}{2}k\end{matrix}\right.\)
Mà \(a+b+c=210\)
\(\Rightarrow\dfrac{7}{6}k+\dfrac{11}{9}k+\dfrac{3}{2}k=210\)
\(\Rightarrow\dfrac{35}{9}k=210\)
\(\Rightarrow k=54\)
\(\Rightarrow\left\{{}\begin{matrix}a=63\\b=66\\c=81\end{matrix}\right.\)
Vậy \(a=63;b=66;c=81\)
Giải:
Ta có: \(\dfrac{6}{7}a=\dfrac{9}{11}b=\dfrac{2}{3}c\Rightarrow\dfrac{a}{\dfrac{7}{6}}=\dfrac{b}{\dfrac{11}{9}}=\dfrac{c}{\dfrac{3}{2}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{\dfrac{7}{6}}=\dfrac{b}{\dfrac{11}{9}}=\dfrac{c}{\dfrac{3}{2}}=\dfrac{210}{\dfrac{35}{9}}=\dfrac{2}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{7}{9}\\b=\dfrac{22}{27}\\c=1\end{matrix}\right.\)
Vậy \(a=\dfrac{7}{9};b=\dfrac{22}{27};c=1\)