d: \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+z}{5+4+3}=\dfrac{48}{12}=4\)
=>x=20; y=16; z=12
\(b,\Leftrightarrow\left[{}\begin{matrix}9-7x=5x-3\left(x\le\dfrac{9}{7}\right)\\7x-9=5x-3\left(x>\dfrac{9}{7}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=3\left(tm\right)\end{matrix}\right.\\ c,\Leftrightarrow\dfrac{12x}{60}=\dfrac{15y}{60}=\dfrac{20z}{60}\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+z}{5+4+3}=\dfrac{48}{12}=4\\ \Leftrightarrow\left\{{}\begin{matrix}x=20\\y=16\\z=12\end{matrix}\right.\)
