\(\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow\dfrac{a}{10}=\dfrac{b}{15};\dfrac{b}{5}=\dfrac{c}{4}\Rightarrow\dfrac{b}{15}=\dfrac{c}{12}.\)
Do đó : \(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}=\dfrac{a-b+c}{10-15+12}=\dfrac{-49}{7}=-7.\)
\(\Rightarrow a=-70;b=-105;c=-84.\)
Theo đề bài: \(\dfrac{a}{2}=\dfrac{b}{3}\); \(\dfrac{b}{5}=\dfrac{c}{4}\)
\(\Rightarrow\) \(\dfrac{a}{10}=\dfrac{b}{15}\); \(\dfrac{b}{15}=\dfrac{c}{12}\)
\(\Rightarrow\) \(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}=\dfrac{a-b+c}{10-15+12}=\dfrac{-49}{7}=-7\)
\(\Rightarrow\dfrac{a}{10}=-7\Rightarrow a=-70\)
và \(\dfrac{b}{15}=-7\Rightarrow b=-105\)
và \(\dfrac{c}{12}=-7\Rightarrow c=-84\)
Vậy \(a=-70\); \(b=-105\); \(c=-84\)
Theo đề bài ta có :
\(\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow\dfrac{a}{10}=\dfrac{b}{15};\dfrac{b}{5}=\dfrac{c}{4}\Rightarrow\dfrac{b}{15}=\dfrac{c}{12}\)
\(\Rightarrow\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}\)
Đặt \(k=\dfrac{a}{10}=\dfrac{b}{15}=\dfrac{c}{12}\)
\(\Rightarrow a=10k;b=15k;c=12k\)
Mà \(a-b+c=-49\) (theo đề bài)
\(\Rightarrow10k-15k+12k=-49\)
\(\left(10-15+12\right).k=-49\)
\(7k=49\)
\(\Rightarrow k=-49\div7=-7\)
\(\Rightarrow a=10k=10.\left(-7\right)=-70\)
\(b=15k=15.\left(-7\right)=-105\)
\(c=12k=12.\left(-7\right)=-84\)
Vậy a = -70 ; b = -105 và c = -84
