Đáp án:
$\begin{array}{l}
7)\dfrac{x}{{0,3}} = \dfrac{y}{{0,2}} = 2z\\
\Leftrightarrow \dfrac{x}{{\dfrac{3}{{10}}}} = \dfrac{y}{{\dfrac{2}{{10}}}} = 2z\\
\Leftrightarrow \dfrac{{10x}}{3} = 5y = 2z\\
\Leftrightarrow \dfrac{1}{{10}}.\dfrac{{10x}}{3} = \dfrac{1}{{10}}.5y = \dfrac{1}{{10}}.2z\\
\Leftrightarrow \dfrac{x}{3} = \dfrac{y}{2} = \dfrac{z}{5} = \dfrac{{3x}}{9} = \dfrac{{z - 3x}}{{5 - 9}} = \dfrac{{ - 1}}{4}\\
\Leftrightarrow x = \dfrac{{ - 3}}{4},y = \dfrac{{ - 1}}{2},z = \dfrac{{ - 5}}{4}\\
Vậy\,x = \dfrac{{ - 3}}{4},y = \dfrac{{ - 1}}{2},z = \dfrac{{ - 5}}{4}\\
8)\dfrac{x}{5} = \dfrac{y}{3} = \dfrac{z}{4} = \dfrac{{2x}}{{10}} = \dfrac{{2x + y - z}}{{10 + 3 - 4}} = \dfrac{{81}}{9} = 9\\
\Leftrightarrow x = 45,y = 27,z = 36\\
Vậy\,x = 45,y = 27,z = 36\\
9)\dfrac{x}{3} = \dfrac{y}{5} = \dfrac{z}{2} = \dfrac{{5x}}{{15}} = \dfrac{{3z}}{6} = \dfrac{{5x - y + 3z}}{{15 - 5 + 6}}\\
= \dfrac{{124}}{{16}} = \dfrac{{31}}{4}\\
\Leftrightarrow x = \dfrac{{93}}{4},y = \dfrac{{155}}{4},z = \dfrac{{31}}{2}\\
Vậy\,x = \dfrac{{93}}{4},y = \dfrac{{155}}{4},z = \dfrac{{31}}{2}\\
10)\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{5} = k\\
\Leftrightarrow x = 2k,y = 3k,z = 5k\\
Khi đó:\\
xyz = 810\\
\Leftrightarrow 2k.3k.5k = 810\\
\Leftrightarrow 30{k^3} = 810\\
\Leftrightarrow {k^3} = 27\\
\Leftrightarrow k = 3\\
\Leftrightarrow x = 6,y = 9,z = 15\\
Vậy\,x = 6,y = 9,z = 15\\
11)\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{6} = k\\
\Leftrightarrow x = 2k,y = 3k,z = 6k\\
Khi đó:\\
{x^2}.{y^2}.{z^2} = {288^2}\\
\Leftrightarrow {\left( {2k} \right)^2}.{\left( {3k} \right)^2}.{\left( {6k} \right)^2} = {288^2}\\
\Leftrightarrow \left[ \begin{array}{l}
2k.3k.6k = 288\\
2k.3k.6k = - 288
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
36{k^3} = 288\\
36{k^3} = - 288
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
{k^3} = 8\\
{k^3} = - 8
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
k = 2\\
k = - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4,y = 6,z = 12\\
x = - 4,y = - 6,z = - 12
\end{array} \right.\\
Vậy\,x = 4,x = - 4,y = 6,y = - 6,z = 12,z = - 12
\end{array}$
