Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x\# 0\\
y\# 0\\
x\# y
\end{array} \right.\\
b)A = \dfrac{2}{x} - \left( {\dfrac{{{x^2}}}{{{x^2} - xy}} + \dfrac{{{x^2} - {y^2}}}{{xy}} - \dfrac{{{y^2}}}{{{y^2} - xy}}} \right):\dfrac{{{x^2} - xy + {y^2}}}{{x - y}}\\
= \dfrac{2}{x} - \left( {\dfrac{{{x^2}y - \left( {{x^2} - {y^2}} \right)\left( {x - y} \right) + {y^2}.x}}{{xy\left( {x - y} \right)}}} \right).\dfrac{{x - y}}{{{x^2} - xy + {y^2}}}\\
= \dfrac{2}{x} - \dfrac{{{x^2}y + x{y^2} - {{\left( {x - y} \right)}^2}\left( {x + y} \right)}}{{xy\left( {x - y} \right)}}.\dfrac{{x - y}}{{{x^2} - xy + {y^2}}}\\
= \dfrac{2}{x} - \dfrac{{\left( {x + y} \right)\left( {xy - {{\left( {x - y} \right)}^2}} \right)}}{{xy}}.\dfrac{1}{{{x^2} - xy + {y^2}}}\\
= \dfrac{2}{x} - \dfrac{{\left( {x + y} \right).\left( {xy - {x^2} + 2xy - {y^2}} \right)}}{{xy\left( {{x^2} - xy + {y^2}} \right)}}\\
= \dfrac{2}{x} - \dfrac{{\left( {x + y} \right).\left( { - {x^2} + xy - {y^2}} \right)}}{{xy\left( {{x^2} - xy + {y^2}} \right)}}\\
= \dfrac{2}{x} + \dfrac{{x + y}}{{xy}}\\
= \dfrac{{2y + x + y}}{{xy}}\\
= \dfrac{{x + 3y}}{{xy}}\\
c)\left| {2x - 1} \right| = 1\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 1\\
2x - 1 = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = 0\left( {ktm} \right)
\end{array} \right.\\
\left| {y + 1} \right| = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
y + 1 = \dfrac{1}{2}\\
y + 1 = - \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
y = - \dfrac{1}{2}\left( {tm} \right)\\
y = - \dfrac{3}{2}\left( {tm} \right)
\end{array} \right.\\
+ Khi:x = 1;y = - \dfrac{1}{2}\\
\Leftrightarrow A = \dfrac{{x + 3y}}{{xy}} = \dfrac{{1 - \dfrac{3}{2}}}{{1.\dfrac{{ - 1}}{2}}} = 1\\
+ Khi:x = 1;y = - \dfrac{3}{2}\\
\Leftrightarrow A = \dfrac{{x + 3y}}{{xy}} = \dfrac{{1 + 3.\dfrac{{ - 3}}{2}}}{{1.\dfrac{{ - 3}}{2}}} = \dfrac{7}{3}\\
d)A = 1\\
\Leftrightarrow \dfrac{{x + 3y}}{{xy}} = 1\\
\Leftrightarrow x + 3y = x.y\\
\Leftrightarrow x - xy + 3y = 0\\
\Leftrightarrow x.\left( {1 - y} \right) + 3y - 3 = - 3\\
\Leftrightarrow x.\left( {1 - y} \right) - 3\left( {1 - y} \right) = - 3\\
\Leftrightarrow \left( {x - 3} \right).\left( {y - 1} \right) = 3 = 1.3 = \left( { - 1} \right).\left( { - 3} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 3 = 1\\
y - 1 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 = 3\\
y - 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 = - 1\\
y - 1 = - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 = - 3\\
y - 1 = - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4;y = 4\left( {ktm} \right)\\
x = 6;y = 2\left( {tm} \right)\\
x = 2;y = - 2\left( {tm} \right)\\
x = 0;y = 0\left( {ktm} \right)
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {6;2} \right);\left( {2; - 2} \right)} \right\}
\end{array}$
