a: Đặt \(\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-3}{5}=k\)
=>\(\left\{{}\begin{matrix}x-1=2k\\y+1=3k\\z-3=5k\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2k+1\\y=3k-1\\z=5k+3\end{matrix}\right.\)
2x+y-z=0
=>\(2\left(2k+1\right)+3k-1-5k-3=0\)
=>4k+2-2k-4=0
=>2k-2=0
=>2k=2
=>k=1
\(x=2\cdot1+1=3;y=3\cdot1-1=3-1=2;z=5\cdot1+3=8\)
b:
\(\dfrac{x}{2}=\dfrac{y}{3}\)
=>\(\dfrac{x}{8}=\dfrac{y}{12}\left(2\right)\)
\(\dfrac{y}{4}=\dfrac{z}{5}\)
=>\(\dfrac{y}{12}=\dfrac{z}{15}\)(1)
Từ (1),(2) suy ra \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Đặt \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=k\)
=>x=8k; y=12k; z=15k
x+y+z=35
=>8k+12k+15k=35
=>35k=35
=>k=1
\(x=8\cdot1=8;y=12\cdot1=12;z=15\cdot1=15\)
