Đáp án:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\begin{array}{l}
b)\dfrac{x}{{ - 2}} = \dfrac{{ - y}}{4} = \dfrac{z}{5}\\
\Leftrightarrow \dfrac{x}{{ - 2}} = \dfrac{y}{{ - 4}} = \dfrac{z}{5} = \dfrac{{2y}}{{ - 8}} = \dfrac{{3z}}{{15}}\\
= \dfrac{{x - 2y + 3z}}{{ - 2 - \left( { - 8} \right) + 15}} = \dfrac{{1200}}{{21}} = \dfrac{{400}}{7}\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{400}}{7}.\left( { - 2} \right) = \dfrac{{ - 800}}{7}\\
y = \dfrac{{400}}{7}.\left( { - 4} \right) = \dfrac{{ - 1600}}{7}\\
z = \dfrac{{400}}{7}.5 = \dfrac{{2000}}{7}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 800}}{7};y = \dfrac{{ - 1600}}{7};z = \dfrac{{2000}}{7}\\
e)\dfrac{x}{{10}} = \dfrac{y}{5} \Leftrightarrow \dfrac{x}{{20}} = \dfrac{y}{{10}}\\
\dfrac{y}{2} = \dfrac{z}{3} \Leftrightarrow \dfrac{y}{{10}} = \dfrac{z}{{15}}\\
\Leftrightarrow \dfrac{x}{{20}} = \dfrac{y}{{10}} = \dfrac{z}{{15}} = \dfrac{{2x}}{{40}} = \dfrac{{3y}}{{30}} = \dfrac{{4z}}{{60}}\\
= \dfrac{{2x - 3y + 4z}}{{40 - 30 + 60}} = \dfrac{{330}}{{70}} = \dfrac{{33}}{7}\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{33}}{7}.20 = \dfrac{{660}}{7}\\
y = \dfrac{{33}}{7}.10 = \dfrac{{330}}{7}\\
z = \dfrac{{33}}{7}.15 = \dfrac{{495}}{7}
\end{array} \right.\\
Vậy\,x = \dfrac{{660}}{7};y = \dfrac{{330}}{7};z = \dfrac{{495}}{7}
\end{array}$
