Đáp án:
$\begin{array}{l}
a)\text{Đặt}:\dfrac{x}{2} = \dfrac{y}{4} = k\\
\Rightarrow \left\{ \begin{array}{l}
x = 2k\\
y = 4k
\end{array} \right.\\
Do:{x^2}.{y^2} = 2\\
\Rightarrow {\left( {2k} \right)^2}.{\left( {4k} \right)^2} = 2\\
\Rightarrow 4{k^2}.16{k^2} = 2\\
\Rightarrow {k^4} = \dfrac{2}{{64}} = \dfrac{1}{{32}}\\
\Rightarrow k = \dfrac{1}{{2\sqrt 2 }}\\
\Rightarrow \left\{ \begin{array}{l}
x = 2.k = \dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}\\
y = 4k = \sqrt 2
\end{array} \right.\\
b)\dfrac{x}{2} = \dfrac{y}{5}
\end{array}$
Theo t/c của dãy tỉ số bằng nhau ta có:
$\begin{array}{l}
\dfrac{x}{2} = \dfrac{y}{5} = \dfrac{{x + y}}{{2 + 5}} = \dfrac{{ - 63}}{7} = - 9\\
\Rightarrow \left\{ \begin{array}{l}
x = \left( { - 9} \right).2 = - 18\\
y = - 45
\end{array} \right.
\end{array}$
c) Theo t/c dãy tỉ số bằng nhau ta có:
$\begin{array}{l}
\dfrac{x}{2} = \dfrac{y}{5} = \dfrac{{x - y}}{{2 - 5}} = \dfrac{{ - 109}}{{ - 3}} = \dfrac{{109}}{3}\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{218}}{3}\\
y = \dfrac{{545}}{3}
\end{array} \right.
\end{array}$
