Question

\(\dfrac{x}{3}=\dfrac{y}{4}\) và x.y = 192

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) và x + y + z = \(-90\)

\(x:y:z=3:8:5\) và 3x + y \(-2z=14\)

Answer 1

1.Tìm x, y.

2.TÌM x, y, z.

3.TÌM x, y, z.

Answer 2

1) \(\dfrac{x}{3}=\dfrac{y}{4}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)

\(\Rightarrow xy=12k^2=192\Rightarrow k=\pm4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm12\\y=\pm16\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=12\\y=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-12\\y=-16\end{matrix}\right.\end{matrix}\right.\)

2) Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{-90}{9}=-10\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-10\right).2=-20\\y=\left(-10\right).3=-30\\z=\left(-10\right).5=-50\end{matrix}\right.\)

3) Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2z}{10}=\dfrac{3x+y-2z}{9+8-10}=\dfrac{14}{7}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.8=16\\z=2.5=10\end{matrix}\right.\)