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$1)$ Điều kiện: $\left\{ \begin{array}{l}
x > 0,\,\,y > 0\\
x \ne 1,\,\,y \ne 1
\end{array} \right.$
$(1)\,\, \Leftrightarrow \,\left\{ \begin{array}{l}
\,x - y = xy\\
x + y = 1
\end{array} \right.\,\,\,\, \Leftrightarrow \,\,\,\,\left\{ \begin{array}{l}
y = 1 - x\\
{x^2} + x - 1 = 0
\end{array} \right.$
$\,\,\,\,\,\,\, \Leftrightarrow \,\,\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = \frac{{ - 1 - \sqrt 5 }}{2} < 0\\
y = \frac{{3 + \sqrt 5 }}{2} > 0
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,(l)\\
\left\{ \begin{array}{l}
x = \frac{{ - 1 + \sqrt 5 }}{2} > 0\\
y = \frac{{3 - \sqrt 5 }}{2} > 0
\end{array} \right.
\end{array} \right.$
$ \Leftrightarrow
$ $\left\{ \begin{array}{l}
x = \frac{{ - 1 + \sqrt 5 }}{2}\\
y = \frac{{3 - \sqrt 5 }}{2}
\end{array} \right.$
$2)$ Ta có: Điều kiện $x > 0$
$ Hệ PT \Leftrightarrow $ $\left\{ \begin{array}{l}
x = {4^{y - 1}}\\
{x^y} = 4096
\end{array} \right.\,\, \Leftrightarrow \,\,\,\left\{ \begin{array}{l}
x = {4^{y - 1}}\\
{4^{y\left( {y - 1} \right)}} = {4^6}
\end{array} \right.$
$ \Leftrightarrow \,\,\,\left\{ \begin{array}{l}
x = {4^{y - 1}}\\
{y^2} - y - 6 = 0
\end{array} \right.$ $ \Leftrightarrow (x;y)= $ $(16,\,3),\,\,\,\left( {\frac{1}{{64}},\, - 2} \right)$
$3)$ Ta có: Điều kiện :$\left\{ \begin{array}{l}
x > 0,\,\,x \ne 1\\
y > 0
\end{array} \right.$
$Hệ (3) \Leftrightarrow
$ $\left\{ \begin{array}{l}
y = {x^2}\\
{x^3} + 2{x^2} + 3x - 22 = 0
\end{array} \right.$
$\Leftrightarrow (x;y)= $ $(2,\,4)$
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