Đáp án:
$\begin{array}{l}
a)\dfrac{x}{y} = \dfrac{3}{4}\\
\Leftrightarrow \dfrac{x}{3} = \dfrac{y}{4} = \dfrac{{2x}}{6} = \dfrac{{5y}}{{20}} = \dfrac{{2x + 5y}}{{6 + 20}} = \dfrac{{10}}{{26}} = \dfrac{5}{{13}}\\
\Leftrightarrow x = \dfrac{{15}}{{13}},y = \dfrac{{20}}{{13}}\\
Vậy\,x = \dfrac{{15}}{{13}},y = \dfrac{{20}}{{13}}\\
b)\dfrac{{2x}}{{3y}} = \dfrac{{ - 1}}{3}\\
\Leftrightarrow \dfrac{{2x}}{{ - 1}} = \dfrac{{3y}}{3} = \dfrac{{2x + 3y}}{{ - 1 + 3}} = \dfrac{7}{2}\\
\Leftrightarrow x = \dfrac{{ - 7}}{4},y = \dfrac{7}{2}\\
Vậy\,x = \dfrac{{ - 7}}{4},y = \dfrac{7}{2}\\
c)21.x = 19.y\\
\Leftrightarrow \dfrac{x}{{19}} = \dfrac{y}{{21}} = \dfrac{{x - y}}{{19 - 21}} = \dfrac{4}{{ - 2}} = - 2\\
\Leftrightarrow x = - 38,y = - 42\\
Vậy\,x = - 38,y = - 42\\
d)Đặt:\dfrac{x}{3} = \dfrac{y}{7} = k\\
\Leftrightarrow x = 3k,y = 7k\\
\Leftrightarrow x.y = 84\\
\Leftrightarrow 3k.7k = 84\\
\Leftrightarrow {k^2} = 4\\
\Leftrightarrow k = \pm 2\\
Khi:k = 2\\
\Leftrightarrow x = 6,y = 14\\
Khi:k = - 2\\
\Leftrightarrow x = - 6,y = - 14\\
Vậy\,\left( {x;y} \right) = \left( {6;14} \right),\left( { - 6; - 14} \right)
\end{array}$
