Đáp án:
a)$x = 4\sqrt 3 ;y = \dfrac{{20}}{{\sqrt 3 }};z = 6\sqrt 3 $ hoặc $x = - 4\sqrt 3 ;y = \dfrac{{ - 20}}{{\sqrt 3 }};z = - 6\sqrt 3 $
b)$ \not \exists x,y$ thỏa mãn đề.
Giải thích các bước giải:
$\begin{array}{l}
a)\dfrac{x}{3} = \dfrac{y}{5} = \dfrac{z}{6}\\
\Rightarrow {\left( {\dfrac{x}{3}} \right)^2} = \dfrac{x}{3}.\dfrac{y}{5} = \dfrac{{xy}}{{3.5}} = \dfrac{{xy}}{{15}} = \dfrac{{80}}{{15}} = \dfrac{{16}}{3}\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{x}{3} = \dfrac{4}{{\sqrt 3 }}\\
\dfrac{x}{3} = - \dfrac{4}{{\sqrt 3 }}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 4\sqrt 3 \\
x = - 4\sqrt 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\sqrt 3 ;y = \dfrac{{20}}{{\sqrt 3 }};z = 6\sqrt 3 \\
x = - 4\sqrt 3 ;y = \dfrac{{ - 20}}{{\sqrt 3 }};z = - 6\sqrt 3
\end{array} \right.
\end{array}$
Vậy $x = 4\sqrt 3 ;y = \dfrac{{20}}{{\sqrt 3 }};z = 6\sqrt 3 $ hoặc $x = - 4\sqrt 3 ;y = \dfrac{{ - 20}}{{\sqrt 3 }};z = - 6\sqrt 3 $
$\begin{array}{l}
b)\dfrac{x}{y} = \dfrac{{17}}{3}\\
\Rightarrow \dfrac{x}{{17}} = \dfrac{y}{3}\\
\Rightarrow {\left( {\dfrac{x}{{17}}} \right)^2} = \dfrac{x}{{17}}.\dfrac{y}{3} = \dfrac{{xy}}{{51}} = \dfrac{{ - 60}}{{51}} = \dfrac{{ - 20}}{{17}}\left( {voly} \right)
\end{array}$
$ \Rightarrow \not \exists x,y$ thỏa mãn đề.
