Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{t}{8}=\dfrac{2y+z}{2\cdot5+6}=\dfrac{-8}{16}=-\dfrac{1}{2}\)
Do đó: x=-3/2; y=-5/2; y=-3; t=-4
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{t}{8}=\dfrac{2y+z}{10+6}=\dfrac{-8}{16}=-\dfrac{1}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-\dfrac{5}{2}\\z=-3\\t=-4\end{matrix}\right.\)
\(x:y:z:t=3:5:6:8\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{t}{8}\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{t}{8}=\dfrac{2y+z}{2.5+6}=\dfrac{-8}{16}=-\dfrac{1}{2}\)
\(\dfrac{x}{3}=-\dfrac{1}{2}\Rightarrow x=-\dfrac{3}{2}\\ \text{}\text{}\dfrac{y}{5}=-\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\ \dfrac{z}{6}=-\dfrac{1}{2}\Rightarrow z=-3\\ \dfrac{t}{8}=-\dfrac{1}{2}\Rightarrow t=-4\)