Ta có: \(3a=7b\Rightarrow\dfrac{a}{7}=\dfrac{b}{3}\)
\(4b=3c\Rightarrow\dfrac{b}{3}=\dfrac{c}{4}\)
Khi đó: \(\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{7+12-20}=\dfrac{-30}{-1}=30\)
Do \(\left\{{}\begin{matrix}\dfrac{a}{7}=30\\\dfrac{4b}{12}=30\\\dfrac{5c}{20}=30\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=210\\b=90\\c=120\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}a=210\\b=90\\c=120\end{matrix}\right.\).
Từ đề ta có: \(\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}\)
Áp dụng t/c của dãy tỉ số = nhau ta có:
\(\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{7+12-20}=\dfrac{-30}{-1}=30\)
\(\Rightarrow\left\{{}\begin{matrix}a=30\cdot7\\b=30\cdot3\\c=30\cdot4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=210\\b=90\\c=120\end{matrix}\right.\)
Vậy..........
\(\text{Theo bài ra ta có:}\)
\(3a=7b\Rightarrow\dfrac{3a}{7}=b\Rightarrow\dfrac{a}{7}=\dfrac{b}{3}\\ 4b=3c\Rightarrow\dfrac{4b}{3}=c\Rightarrow\dfrac{b}{3}=\dfrac{c}{4}\\ \Rightarrow\dfrac{a}{7}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}\)
\(a+4b-5c=-30\)
\(\text{Áp dụng tính chất dãy tỉ số bằng nhau ta được : }\)
\(\dfrac{a}{7}=\dfrac{4b}{12}=\dfrac{5c}{20}=\dfrac{a+4b-5c}{7+12-20}=\dfrac{-30}{-1}=30\) \(\left(1\right)\)
\(\text{Từ}\) \(\left(1\right)\) \(\text{suy ra : }\) \(\left\{{}\begin{matrix}\dfrac{a}{7}=30\Rightarrow a=210\\\dfrac{b}{3}=30\Rightarrow b=90\\\dfrac{c}{4}=30\Rightarrow c=120\end{matrix}\right.\)
\(\text{Vậy}\) \(a=210\\ b=90\\ c=120\)
\(\)