a) $\dfrac{x}{10} = \dfrac{y}{6} = \dfrac{z}{21}$

⇔ $\dfrac{5x}{50} = \dfrac{y}{6} = \dfrac{-2z}{-42}$

⇔ $\dfrac{5x}{50} = \dfrac{y}{6} = \dfrac{-2z}{-42} = \dfrac{5x + y -2z}{50 + 6 – 42} = \dfrac{28}{14} = 2$

⇔ $\dfrac{x}{10} = \dfrac{y}{6} = \dfrac{z}{21} = 2$

⇒ $\begin{cases} \dfrac{x}{10} = 2 \Rightarrow x = 10. 2 =20\\\dfrac{y}{6} = 2 \Rightarrow y = 6. 2 =12\\\dfrac{z}{21} = 2 \Rightarrow z = 21.2 = 42\end{cases}$

b) $3x = 2y; \, 7y = 5z$
⇔ $\left \{ {{x = \dfrac{2y}{3}} \atop {z = \dfrac{7y}{5}}} \right.$ 

Ta có: $x – y + z = 32$

⇒ $\dfrac{2y}{3} – y + \dfrac{7y}{5} = 32$

⇒ $\dfrac{16y}{15} = 32$

⇒ $y = 30 ⇒ x = \dfrac{2.30}{3} = 20 ⇒ z = \dfrac{7.30}{5} = 42$

c) $\dfrac{x – 1}{2} = \dfrac{y – 2}{3} = \dfrac{z – 3}{4}$

⇔ $\dfrac{2(x – 1)}{4} = \dfrac{3(y – 2)}{9} = \dfrac{-(z – 3)}{-4}$

⇔ $\dfrac{2x – 2 }{4} = \dfrac{3y – 6}{9} = \dfrac{-z + 3}{-4}$

⇔ $\dfrac{2x – 2}{4} = \dfrac{3y – 6}{9} = \dfrac{-z + 3}{-4} = \dfrac{(2x + 3y – z) – 2 – 6 + 3}{4 + 9 – 4} = \dfrac{45}{9} = 5$

⇔ $\dfrac{x – 1}{2} = \dfrac{y – 2}{3} = \dfrac{z – 3}{4} = 5$

$\begin{cases}\dfrac{x – 1}{2} = 5 \Leftrightarrow x – 1 = 10 \Leftrightarrow x = 11 \\\dfrac{y – 2}{3} = 5 \Leftrightarrow y – 2 = 15 \Leftrightarrow y = 17\\\dfrac{z – 3}{4} = 5 \Leftrightarrow z – 3 = 20 \Leftrightarrow z = 23  \end{cases}$

d) $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{5}$

⇔ $\begin{cases}x = \dfrac{2y}{3}\\z = \dfrac{5y}{3}\end{cases}$

Ta có: $xyz = 810$

⇒ $\dfrac{2y}{3}\cdot y \cdot \dfrac{5y}{3} = 810$

⇒ $\dfrac{10y^{3}}{9} = 810$

⇒ $y^{3} = 729$

⇒ $y = \sqrt[3]{729} = 9 ⇒ x = \dfrac{2.9}{3} = 6 ⇒ z = \dfrac{5.9}{3} = 15$