Đáp án:
\(\begin{array}{l}
*)\\
\left[ \begin{array}{l}
x = 1,\,\,\forall y\\
y = 1,\,\,\,\,\forall x
\end{array} \right.\\
*)\\
\left[ \begin{array}{l}
x = 5\\
x = \sqrt 3 \\
x = - \sqrt 3
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
x + y = xy + 1\\
\Leftrightarrow xy + 1 - \left( {x + y} \right) = 0\\
\Leftrightarrow xy + 1 - x - y = 0\\
\Leftrightarrow \left( {xy - x} \right) + \left( { - y + 1} \right) = 0\\
\Leftrightarrow x.\left( {y - 1} \right) - \left( {y - 1} \right) = 0\\
\Leftrightarrow \left( {y - 1} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
y - 1 = 0\\
x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1,\,\,\forall y\\
y = 1,\,\,\,\,\forall x
\end{array} \right.\\
*)\\
{x^2}\left( {x - 5} \right) + 15 - 3x = 0\\
\Leftrightarrow {x^2}\left( {x - 5} \right) - \left( {3x - 15} \right) = 0\\
\Leftrightarrow {x^2}\left( {x - 5} \right) - 3\left( {x - 5} \right) = 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {{x^2} - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
{x^2} - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
{x^2} = 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \sqrt 3 \\
x = - \sqrt 3
\end{array} \right.
\end{array}\)
