Đáp án:
$\begin{array}{l}
a)\dfrac{x}{{13}} = \dfrac{y}{4} = \dfrac{{x - y}}{{13 - 4}} = \dfrac{{18}}{9} = 2\\
\Rightarrow \left\{ \begin{array}{l}
x = 13.2 = 26\\
y = 4.2 = 8
\end{array} \right.\\
c)\dfrac{x}{3} = \dfrac{y}{4} = \dfrac{{y - x}}{{4 - 3}} = \dfrac{{ - 5}}{1} = - 5\\
\Rightarrow \left\{ \begin{array}{l}
x = - 15\\
y = - 20
\end{array} \right.\\
b)\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{{x + y}}{{2 + 3}} = \dfrac{{ - 15}}{5} = - 3\\
\Rightarrow \left\{ \begin{array}{l}
x = - 6\\
y = - 9
\end{array} \right.\\
d)\dfrac{x}{9} = \dfrac{y}{{ - 5}} = \dfrac{{x + y}}{{9 - 5}} = \dfrac{{ - 2}}{4} = - \dfrac{1}{2}\\
\Rightarrow \left\{ \begin{array}{l}
x = \dfrac{{ - 9}}{2}\\
y = - \dfrac{5}{2}
\end{array} \right.\\
e)\dfrac{x}{y} = \dfrac{4}{9}\\
\Rightarrow \dfrac{x}{4} = \dfrac{y}{9} = \dfrac{{3x}}{{12}} = \dfrac{{2y}}{{18}} = \dfrac{{3x - 2y}}{{12 - 18}} = \dfrac{{ - 12}}{{ - 6}} = 2\\
\Rightarrow \left\{ \begin{array}{l}
x = 8\\
y = 18
\end{array} \right.\\
f)\dfrac{y}{4} = \dfrac{x}{{ - 3}} = \dfrac{{x - y}}{{ - 3 - 4}} = \dfrac{7}{{ - 7}} = - 1\\
\Rightarrow \left\{ \begin{array}{l}
x = 3\\
y = - 4
\end{array} \right.\\
g)\dfrac{1}{3}x = \dfrac{{ - 1}}{7}y\\
\Rightarrow \dfrac{x}{3} = \dfrac{{ - y}}{7} = \dfrac{{x - y}}{{3 + 7}} = \dfrac{{ - 20}}{{10}} = - 2\\
\Rightarrow \left\{ \begin{array}{l}
x = - 6\\
y = 14
\end{array} \right.\\
h)\dfrac{x}{y} = 3,5 = \dfrac{7}{2}\\
\Rightarrow \dfrac{x}{7} = \dfrac{y}{2} = \dfrac{{x - y}}{{7 - 2}} = \dfrac{{10}}{5} = 2\\
\Rightarrow \left\{ \begin{array}{l}
x = 14\\
y = 4
\end{array} \right.\\
i)\dfrac{x}{y} = \dfrac{7}{{13}}\\
\Rightarrow \dfrac{x}{7} = \dfrac{y}{{13}} = k \Rightarrow \left\{ \begin{array}{l}
x = 7k\\
y = 13k
\end{array} \right.\\
\Rightarrow x.y = 7k.13k = 91\\
\Rightarrow {k^2} = 1\\
\Rightarrow \left[ \begin{array}{l}
k = 1 \Rightarrow x = 7;y = 13\\
k = - 1 \Rightarrow x = - 7;y = - 13
\end{array} \right.\\
j)x = - 2y\\
\Rightarrow \dfrac{x}{{ - 2}} = \dfrac{y}{1} = \dfrac{{x - y}}{{ - 2 - 1}} = \dfrac{{ - 3}}{{ - 3}} = 1\\
\Rightarrow \left\{ \begin{array}{l}
x = - 2\\
y = 1
\end{array} \right.\\
k)\dfrac{x}{2} = \dfrac{y}{5} = \dfrac{z}{7} = \dfrac{{2x}}{4} = \dfrac{{2x + y - z}}{{4 + 5 - 7}} = \dfrac{2}{2} = 1\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 5\\
z = 7
\end{array} \right.\\
m)2x = 3y = 4z\\
\Rightarrow \dfrac{{2x}}{{12}} = \dfrac{{3y}}{{12}} = \dfrac{{4z}}{{12}}\\
\Rightarrow \dfrac{x}{6} = \dfrac{y}{4} = \dfrac{z}{3} = \dfrac{{x + y - z}}{{6 + 4 - 3}} = \dfrac{{21}}{7} = 3\\
\Rightarrow \left\{ \begin{array}{l}
x = 18\\
y = 12\\
z = 9
\end{array} \right.
\end{array}$
$\begin{array}{l}
n)x = 2y;3y = 4z\\
\Rightarrow \left\{ \begin{array}{l}
y = \dfrac{x}{2}\\
y = \dfrac{4}{3}z
\end{array} \right.\\
\Rightarrow \dfrac{x}{2} = y = \dfrac{4}{3}z\\
\Rightarrow \dfrac{x}{8} = \dfrac{y}{4} = \dfrac{z}{3} = \dfrac{{x + y + z}}{{8 + 4 + 3}} = \dfrac{{60}}{{15}} = 4\\
\Rightarrow \left\{ \begin{array}{l}
x = 32\\
y = 16\\
z = 12
\end{array} \right.\\
o)x:y:z = 2:\left( { - 3} \right):4\\
\Rightarrow \dfrac{x}{2} = \dfrac{y}{{ - 3}} = \dfrac{z}{4} = \dfrac{{x - y}}{{2 + 3}} = \dfrac{{20}}{5} = 4\\
\Rightarrow \left\{ \begin{array}{l}
x = 8\\
y = - 12\\
z = 16
\end{array} \right.
\end{array}$
