Đáp án:
$(x;y) = \left\{(4;-4),(1;4)\right\}$
Giải thích các bước giải:
$\dfrac{x}{2} + \dfrac{3}{y} = \dfrac{5}{4} \qquad (y \ne 0)$
$\Leftrightarrow 4(xy+6) = 5.2y$
$\Leftrightarrow 2xy + 12 = 5y$
$\Leftrightarrow y(2x - 5) = - 12$
Ta có:
$-12 = (-1).12 = (-2).6 = (-3).4 = (-4).3 = (-6). 2= 1.(-12) = 2.(-6) = 3.(-4) = 4.(-3) = 6.(-2)$
Ta có bảng giá trị:
$\begin{array}{|l|r|}
\hline
\quad y & -1\,\,&-2&-3&-4&-6&6&4&3\,&2\,\,&1\,\,\,\\
\hline
2x - 5 &12\,\,&6&4&3&2&-2&-3&-4&-6\,&-12\,\, \\
\hline
\quad x&\dfrac{17}{2}&\dfrac{11}{2}&\dfrac{9}{2}&4&\dfrac{7}{2}&\dfrac{3}{2}&1&\dfrac{1}{2}&-\dfrac{1}{2}&-\dfrac{7}{2}\,\,\\
(x \in \Bbb Z)&(l)&(l)&(l)&(n)&(l)&(l)&(n)&(l)&(l)&(l)\,\,\,\\
\hline
\end{array}$
Vậy $(x;y) = \left\{(4;-4),(1;4)\right\}$
