Question

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4},a+2b-c3=-20\)

Tìm a,b,c

Answer 1

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

\(\Leftrightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}\)

Áp dụng t.c dãy tỉ số bằng nhau ta có :

\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b-3c}{2+6-12}=\dfrac{-20}{-4}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=5\\\dfrac{2b}{6}=5\\\dfrac{3c}{12}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=10\\b=15\\c=20\end{matrix}\right.\)

Vậy ..

Answer 2

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+b+c}{2+2.3+4.3}=\dfrac{-20}{-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=5\Rightarrow a=10\\\dfrac{b}{3}=5\Rightarrow b=15\\\dfrac{c}{4}=5\Rightarrow c=20\end{matrix}\right.\)

Vậy ..................