Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
2{x^2} - 5 = \left( {2{x^2} + 4x} \right) - \left( {4x + 8} \right) + 3 = 2x\left( {x + 2} \right) - 4\left( {x + 2} \right) + 3 = \left( {2x - 4} \right)\left( {x + 2} \right) + 3\\
\left( {2{x^2} - 5} \right) \vdots \left( {x + 2} \right) \Leftrightarrow \left( {2x - 4} \right)\left( {x + 2} \right) + 3 \vdots \left( {x + 2} \right)\\
\Rightarrow 3 \vdots \left( {x + 2} \right)\\
x \in Z \Rightarrow x + 2 \in \left\{ { - 3;\,\, - 1;\,\,1;\,\,3} \right\}\\
\Rightarrow x \in \left\{ { - 5;\,\, - 3;\,\, - 1;\,\,1} \right\}\\
b,\\
5{y^2} - x{y^2} + 17 = 0\\
\Leftrightarrow x{y^2} - 5{y^2} = 17\\
\Leftrightarrow {y^2}\left( {x - 5} \right) = 17\\
{y^2} \ge 0,\,\,\,\,\forall y\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{y^2} = 1\\
x - 5 = 17
\end{array} \right.\\
\left\{ \begin{array}{l}
{y^2} = 17\\
x - 5 = 1
\end{array} \right.\left( {L,\,\,\,y \in Z} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 22\\
y = \pm 1
\end{array} \right.
\end{array}\)
