Đáp án: $\left\{ {\left( {12;\dfrac{{12}}{5}} \right);\left( { - \dfrac{4}{5};\dfrac{4}{7}} \right);\left( {\dfrac{6}{5};\dfrac{6}{7}} \right);\left( { - 2;\dfrac{2}{5}} \right)} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x;y\# 0;x\# y;x\# - y\\
Dat:\left\{ \begin{array}{l}
\dfrac{{x + y}}{{xy}} = \dfrac{1}{x} + \dfrac{1}{y} = a\\
\dfrac{{x - y}}{{xy}} = \dfrac{1}{y} - \dfrac{1}{x} = b
\end{array} \right.\\
\Leftrightarrow \dfrac{{xy}}{{x + y}} = \dfrac{1}{a};\dfrac{{xy}}{{x - y}} = \dfrac{1}{b}\\
\Leftrightarrow \left\{ \begin{array}{l}
a + \dfrac{1}{a} = \dfrac{5}{2}\\
b + \dfrac{1}{b} = \dfrac{{10}}{3}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{a^2} - \dfrac{5}{2}.a + 1 = 0\\
{b^2} - \dfrac{{10}}{3}b + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2{a^2} - 5a + 2 = 0\\
3{b^2} - 10b + 3 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {2a - 1} \right)\left( {a - 2} \right) = 0\\
\left( {3b - 1} \right)\left( {b - 3} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
a = \dfrac{1}{2}\\
a = 2
\end{array} \right.\\
\left[ \begin{array}{l}
b = \dfrac{1}{3}\\
b = 3
\end{array} \right.
\end{array} \right.\\
+ TH1:a = \dfrac{1}{2};b = \dfrac{1}{3}\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{2}\\
\dfrac{1}{y} - \dfrac{1}{x} = \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{x} = \dfrac{1}{{12}}\\
\dfrac{1}{y} = \dfrac{5}{{12}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 12\left( {tm} \right)\\
y = \dfrac{{12}}{5}\left( {tm} \right)
\end{array} \right.\\
+ Khi:a = \dfrac{1}{2};b = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{2}\\
\dfrac{1}{y} - \dfrac{1}{x} = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{x} = - \dfrac{5}{4}\\
\dfrac{1}{y} = \dfrac{7}{4}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - \dfrac{4}{5}\\
y = \dfrac{4}{7}
\end{array} \right.\left( {tm} \right)\\
+ TH2:a = 2;b = \dfrac{1}{3}\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{x} + \dfrac{1}{y} = 2\\
\dfrac{1}{y} - \dfrac{1}{x} = \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{6}{5}\\
y = \dfrac{6}{7}
\end{array} \right.\left( {tmdk} \right)\\
+ TH4:x = 2;y = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{x} + \dfrac{1}{y} = 2\\
\dfrac{1}{y} - \dfrac{1}{x} = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 2\\
y = \dfrac{2}{5}
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {12;\dfrac{{12}}{5}} \right);\left( { - \dfrac{4}{5};\dfrac{4}{7}} \right);\left( {\dfrac{6}{5};\dfrac{6}{7}} \right);\left( { - 2;\dfrac{2}{5}} \right)} \right\}
\end{array}$
