Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {x - 1} \right)\left( {y + 1} \right) = 3\\
3 = 1.3 = \left( { - 1} \right).\left( { - 3} \right)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 = 1\\
y + 1 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 = 3\\
y + 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 = - 1\\
y + 1 = - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 = - 3\\
y + 1 = - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 2\\
y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 4\\
y = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 0\\
y = - 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 2\\
y = - 2
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( {2;2} \right);\left( {4;0} \right);\left( {0; - 4} \right);\left( { - 2; - 2} \right)} \right\}\\
b,\\
x\left( {y + 2} \right) = - 8\\
- 8 = 1.\left( { - 8} \right) = \left( { - 8} \right).1 = 2.\left( { - 4} \right) = 4.\left( { - 2} \right)\\
{T^2}a.\\
c,\\
xy - 2x - 2y = 0\\
\Leftrightarrow \left( {xy - 2x} \right) - \left( {2y - 4} \right) = 4\\
\Leftrightarrow x\left( {y - 2} \right) - 2\left( {y - 2} \right) = 4\\
\Leftrightarrow \left( {x - 2} \right)\left( {y - 2} \right) = 4\\
4 = 1.4 = 2.2 = \left( { - 1} \right).\left( { - 4} \right) = \left( { - 2} \right).\left( { - 2} \right)
\end{array}\)
