b: Ta có: 5x=2y
nên \(\dfrac{x}{2}=\dfrac{y}{5}\)
hay \(\dfrac{x}{6}=\dfrac{y}{15}\left(1\right)\)
Ta có: 2x=3z
nên \(\dfrac{x}{3}=\dfrac{z}{2}\)
hay \(\dfrac{x}{6}=\dfrac{z}{4}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\) suy ra \(\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{4}\)
Đặt \(\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=6k\\y=15k\\z=4k\end{matrix}\right.\)
Ta có: xy=90
\(\Leftrightarrow k^2=1\)
Trường hợp 1: k=1
\(\Leftrightarrow\left\{{}\begin{matrix}x=6k=6\\y=15k=15\\z=4k=4\end{matrix}\right.\)
Trường hợp 2: k=-1
\(\Leftrightarrow\left\{{}\begin{matrix}x=6k=-6\\y=15k=-15\\z=4k=-4\end{matrix}\right.\)
