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: $xyz = 810$
⇒ $\dfrac{2y}{3}\cdot y \cdot \dfrac{5y}{3} = 810$
⇒ $\dfrac{10y^{3}}{9} = 810$
⇒ $y^{3} = 729$
⇒ $y = \sqrt[3]{729} = 9 ⇒ x = \dfrac{2.9}{3} = 6 ⇒ z = \dfrac{5.9}{3} = 15$
