\(\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}\) (1)
\(\dfrac{b}{4}=\dfrac{c}{5}\Rightarrow\dfrac{b}{12}=\dfrac{c}{15}\) (2)
Từ (1) và (2) ta có: \(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{2c}{30}=\dfrac{a+b-2c}{8+12-30}=\dfrac{10}{-10}=-1\)
Vậy \(\left\{{}\begin{matrix}a=-8\\b=-12\\c=-15\end{matrix}\right.\)
\(\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}\left(1\right)\)
\(\dfrac{b}{4}=\dfrac{c}{5}\Rightarrow\dfrac{b}{12}=\dfrac{c}{15}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{2c}{15}=\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{30}\)
Theo bài ra ta có:
\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{30}\) và \(a+b-2c=10\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{2c}{30}=\dfrac{a+b-2c}{8+12-30}=\dfrac{10}{-10}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{8}=-1\Rightarrow a=-8\\\dfrac{b}{12}=-1\Rightarrow b=-12\\\dfrac{c}{15}=-1\Rightarrow c=-15\end{matrix}\right.\)
Chúc bạn học tốt!
