Đáp án:
$\begin{array}{l}
1)\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4} = \dfrac{{2x}}{4} = \dfrac{{3y}}{9} = \dfrac{{5z}}{{20}}\\
= \dfrac{{2x + 3y - 5z}}{{4 + 9 - 20}} = \dfrac{{ - 21}}{{ - 7}} = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 3.2 = 6\\
y = 3.3 = 9\\
z = 3.4 = 12
\end{array} \right.\\
Vậy\,x = 6;y = 9;z = 12\\
2)\\
\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4} = k\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 2k\\
y = 3k\\
z = 4k
\end{array} \right.\\
{x^3} + {y^3} + {z^3} = 792\\
\Leftrightarrow {\left( {2k} \right)^3} + {\left( {3k} \right)^3} + {\left( {4k} \right)^3} = 792\\
\Leftrightarrow 8{k^3} + 27{k^3} + 64{k^3} = 792\\
\Leftrightarrow 99{k^3} = 792\\
\Leftrightarrow {k^3} = 8\\
\Leftrightarrow k = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = 6\\
z = 8
\end{array} \right.\\
Vậy\,x = 4;y = 6;z = 8\\
3)\dfrac{{x - 1}}{2} = \dfrac{{y - 2}}{3} = \dfrac{{z - 3}}{4}\\
= \dfrac{{2\left( {x - 1} \right)}}{4} = \dfrac{{3\left( {y - 2} \right)}}{9}\\
= \dfrac{{2x - 2}}{4} = \dfrac{{3y - 6}}{9}\\
= \dfrac{{2x - 2 + 3y - 6 - \left( {z - 3} \right)}}{{4 + 9 - 4}}\\
= \dfrac{{2x + 3y - z - 2 - 6 + 3}}{9}\\
= \dfrac{{50 - 5}}{9} = 5\\
\Leftrightarrow \left\{ \begin{array}{l}
x - 1 = 10\\
y - 2 = 15\\
z - 3 = 20
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 11\\
y = 17\\
z = 23
\end{array} \right.\\
Vậy\,x = 11;y = 17;z = 23
\end{array}$
