Question

a/2=b/3và a+b=-15

a/2=b/3=c/4 và a+2b-3c=-20

Answer 1

a) Áp dụng TCDTSBN ta có:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=-\dfrac{15}{5}=-3\)

\(\dfrac{a}{2}=-3\Rightarrow a=-6\\ \dfrac{b}{3}=-3\Rightarrow b=-9\)

b) Áp dụng TCDTSBN ta có:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{a+2b-3c}{2+2.3-3.4}=\dfrac{-20}{-4}=5\)

\(\dfrac{a}{2}=5\Rightarrow a=10\\ \dfrac{b}{3}=5\Rightarrow b=15\\ \dfrac{c}{4}=5\Rightarrow c=20\)

 

Answer 2

1) Áp dụng t/c dtsbn:

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{-15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}a=\left(-3\right).2=-6\\b=\left(-3\right).3=-9\end{matrix}\right.\)

2) Áp dụng t/c dtsbn:

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

\(\Rightarrow\left\{{}\begin{matrix}a=5.2=10\\b=5.3=15\\c=5.4=20\end{matrix}\right.\)