Đáp án:
$\begin{array}{l}
c)5x = 3y;4y = 2z\\
\Rightarrow \left\{ \begin{array}{l}
y = \dfrac{{5x}}{3}\\
y = \dfrac{{2z}}{4} = \dfrac{z}{2}
\end{array} \right.\\
\Rightarrow \dfrac{{5x}}{3} = y = \dfrac{z}{2}\\
\Rightarrow \dfrac{{5x}}{{3.5}} = \dfrac{y}{5} = \dfrac{z}{{2.5}}\\
\Rightarrow \dfrac{x}{3} = \dfrac{y}{5} = \dfrac{z}{{10}} = \dfrac{{x + y + z}}{{3 + 5 + 10}} = \dfrac{{48}}{{18}} = \dfrac{8}{3}\\
\Rightarrow \left\{ \begin{array}{l}
x = 8\\
y = \dfrac{{40}}{3}\\
z = \dfrac{{80}}{3}
\end{array} \right.\\
d)\dfrac{x}{3} = \dfrac{y}{5} = k\\
\Rightarrow \left\{ \begin{array}{l}
x = 3k\\
y = 5k
\end{array} \right.\\
Do:{x^2} - {y^2} = - 4\\
\Rightarrow {\left( {3k} \right)^2} - {\left( {5k} \right)^2} = - 4\\
\Rightarrow 9{k^2} - 25{k^2} = - 4\\
\Rightarrow - 16{k^2} = - 4\\
\Rightarrow {k^2} = \dfrac{1}{4}\\
\Rightarrow \left[ \begin{array}{l}
k = \dfrac{1}{2}\\
k = - \dfrac{1}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2};y = \dfrac{5}{2}\\
x = - \dfrac{3}{2};y = - \dfrac{5}{2}
\end{array} \right.
\end{array}$
