a. $\dfrac{x}{0,3} = \dfrac{y}{0,7} = z \to \dfrac{x}{0,3} = \dfrac{y}{0,7} = \dfrac{z}{1} = \dfrac{3x}{0.9} = \dfrac{z - 3x}{1 - 0,9} = \dfrac{1}{0,1} = 10$
Suy ra:
$\dfrac{x}{0,3} = 10 \to x = 0,3.10 = 3$
$\dfrac{y}{0,7} = 10 \to y = 0,7.10 = 7$
$\dfrac{z}{1} = 10 \to z = 1.10 = 10$
b. $\dfrac{x - 5}{3} = \dfrac{y - 4}{4} = \dfrac{z - 3}{5} = \dfrac{x - 5 + y - 4 + z - 3}{3 + 4 + 5} = \dfrac{(x + y + z) - (5 + 4 + 3)}{12} = \dfrac{36 - 12}{12} = \dfrac{24}{12} = 2$
Suy ra:
$\dfrac{x - 5}{3} = 2 \to x - 5 = 3.2 \to x = 6 + 5 = 11$
$\dfrac{y - 4}{4} = 2 \to y - 4 = 4.2 \to y = 8 + 4 = 12$
$\dfrac{z - 3}{5} = 2 \to z - 3 = 5.2 \to z = 10 + 3 = 13$
c. $\dfrac{x - 1}{3} = \dfrac{y - 2}{4} = \dfrac{z + 7}{5} = \dfrac{x - 1 + y - 2 - z - 7}{3 + 4 - 5} = \dfrac{x + y - z - 1 - 2 - 7}{3 + 4 - 5} = \dfrac{8 - 10}{2} = - 1$
Suy ra:
$\dfrac{x - 1}{3} = - 1 \to x - 1 = - 3 \to x = - 2$
$\dfrac{y - 2}{4} = - 1 \to y - 2 = - 4 \to y = - 2$
$\dfrac{z + 7}{5} = - 1 \to z + 7 = - 5 \to z = - 12$
