Question

Tìm x,y,z biết x/3=y/4=z/6 và x+2y-3z=-14

Giúp em câu này vs ạ

 

Answer 1

Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x}{3}=\dfrac{2y}{2.4}=\dfrac{3z}{3.6}\)

Áp dung tcdtsbn , ta có: 

\(\dfrac{x}{3}=\dfrac{2y}{2.4}=\dfrac{3z}{3.6}=\dfrac{x+2y-3z}{3+8-18}=\dfrac{-14}{-7}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=8\\z=12\end{matrix}\right.\)

Answer 2

áp dụng t/c dtsbn ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x+2y-3z}{3+2.4-3.6}=\dfrac{-14}{-7}=2\)

\(\dfrac{x}{3}=2\Rightarrow x=6\\ \dfrac{y}{4}=2\Rightarrow y=8\\ \dfrac{z}{6}=2\Rightarrow z=12\)