Đáp án:
$\begin{array}{l}
1)\dfrac{x}{y} = \dfrac{{ - 7}}{4}\\
\Rightarrow \dfrac{x}{{ - 7}} = \dfrac{y}{4} = \dfrac{{4x}}{{ - 28}} = \dfrac{{5y}}{{20}}\\
= \dfrac{{4x - 5y}}{{ - 28 - 20}} = \dfrac{{72}}{{ - 48}} = \dfrac{{ - 3}}{2}\\
\Rightarrow \left\{ \begin{array}{l}
x = - \dfrac{3}{2}.\left( { - 7} \right) = \dfrac{{21}}{2}\\
y = - \dfrac{3}{2}.4 = - 6
\end{array} \right.\\
2)\dfrac{{x - 1}}{5} = \dfrac{{y - 2}}{3} = \dfrac{{z - 2}}{2}\\
= \dfrac{{2y - 4}}{6}\\
= \dfrac{{x - 1 + 2y - 4}}{{5 + 6}}\\
= \dfrac{{x + 2y - 2 - 3}}{{11}}\\
= \dfrac{{6 - 3}}{{11}} = \dfrac{3}{{11}}\\
\Rightarrow \left\{ \begin{array}{l}
x - 1 = \dfrac{3}{{11}}.5 = \dfrac{{15}}{{11}}\\
y - 2 = \dfrac{3}{{11}}.3 = \dfrac{9}{{11}}\\
z - 2 = \dfrac{3}{{11}}.2 = \dfrac{6}{{11}}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{26}}{{11}}\\
y = \dfrac{{31}}{{11}}\\
z = \dfrac{{28}}{{11}}
\end{array} \right.\\
3)\dfrac{{x - 3}}{{ - 4}} = \dfrac{{y + 4}}{7} = \dfrac{{z - 5}}{3}\\
= \dfrac{{2x - 6}}{{ - 8}} = \dfrac{{4y + 16}}{{28}} = \dfrac{{5z - 25}}{{15}}\\
= \dfrac{{ - 2x + 6 - 4y - 16 + 5z - 25}}{{8 - 28 + 15}}\\
= \dfrac{{ - 2x - 4y + 5z - 35}}{{ - 5}}\\
= \dfrac{{ - 48 - 35}}{{ - 5}} = \dfrac{{83}}{5}\\
\Rightarrow \left\{ \begin{array}{l}
x - 3 = \dfrac{{83}}{5}.\left( { - 4} \right) = \dfrac{{ - 332}}{5}\\
y + 4 = \dfrac{{83}}{5}.7 = \dfrac{{581}}{5}\\
z - 5 = \dfrac{{83}}{5}.3 = \dfrac{{249}}{5}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{ - 317}}{5}\\
y = \dfrac{{561}}{5}\\
z = \dfrac{{274}}{5}
\end{array} \right.\\
4)\dfrac{x}{5} = \dfrac{y}{{ - 4}} = \dfrac{z}{6} = k\\
\Rightarrow \left\{ \begin{array}{l}
x = 5k\\
y = - 4k\\
z = 6k
\end{array} \right.\\
Do:x - y.z = 15\\
\Rightarrow 5k - \left( { - 4k} \right).6k = 15\\
\Rightarrow 5k + 24{k^2} = 15\\
\Rightarrow 24{k^2} + 5k - 15 = 0\\
\Rightarrow \left[ \begin{array}{l}
k = 0,69\\
k = - 0,9
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3,45;y = - 2,76;z = 4,14\\
x = - 4,5;y = 3,6;z = - 5,4
\end{array} \right.
\end{array}$
