Đáp án:
$\begin{array}{l}
3)d)\frac{{x + 1}}{3} = \frac{1}{y}\\
\Rightarrow \left( {x + 1} \right).y = 3 = 3.1 = \left( { - 3} \right).\left( { - 1} \right)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 = 3\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = 1\\
y = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = - 3\\
y = - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 = - 1\\
y = - 3
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 2\\
y = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 0\\
y = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 4\\
y = - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 2\\
y = - 3
\end{array} \right.
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {2;1} \right);\left( {0;3} \right);\left( { - 4; - 1} \right);\left( { - 2; - 3} \right)} \right\}\\
4)\\
\frac{{a + b}}{b} = \frac{a}{b} + \frac{b}{b} = \frac{a}{b} + 1\\
\frac{{c + d}}{d} = \frac{c}{d} + \frac{d}{d} = \frac{c}{d} + 1\\
Do:\frac{a}{b} = \frac{c}{d}\\
\Rightarrow \frac{a}{b} + 1 = \frac{c}{d} + 1\\
\Rightarrow \frac{{a + b}}{b} = \frac{{c + d}}{d}
\end{array}$
