Question

Answer 1

x/3=y/4=k

x=3k,y=4k

3k.4k=48

12k^2=48 

k^2=4

k=+-2

 

Answer 2

2) Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{7}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=7k\end{matrix}\right.\)

Ta có: xy=48

\(\Leftrightarrow12k^2=48\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=3\cdot2=6\\y=4k=4\cdot2=8\\z=7k=7\cdot2=-14\end{matrix}\right.\)

Trường hợp 2: k=-2

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k=3\cdot\left(-2\right)=-6\\y=4k=4\cdot\left(-2\right)=-8\\z=7k=7\cdot\left(-2\right)=-14\end{matrix}\right.\)