Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
\frac{2}{3}x\left( {{x^2} - 4} \right) = 0\\
\Leftrightarrow \frac{2}{3}x\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = - 2
\end{array} \right.\\
b,\\
{\left( {x + 2} \right)^2} - \left( {x - 2} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left( {{x^2} + 2.x.2 + {2^2}} \right) - \left( {{x^2} + 3x - 2x - 6} \right) = 0\\
\Leftrightarrow \left( {{x^2} + 4x + 4} \right) - \left( {{x^2} + x - 6} \right) = 0\\
\Leftrightarrow {x^2} + 4x + 4 - {x^2} - x + 6 = 0\\
\Leftrightarrow 3x + 10 = 0\\
\Leftrightarrow x = - \frac{{10}}{3}\\
2,\\
2{n^2} - n + 2\\
= \left( {2{n^2} + n} \right) + \left( { - 2n - 1} \right) + 3\\
= n\left( {2n + 1} \right) - 1.\left( {2n + 1} \right) + 3\\
= \left( {2n + 1} \right)\left( {n - 1} \right) + 3\\
\left( {2{n^2} - n + 2} \right)\,\, \vdots \,\,\left( {2n + 1} \right)\\
\Leftrightarrow \left[ {\left( {2n + 1} \right)\left( {n - 1} \right) + 3} \right]\,\, \vdots \,\,\left( {2n + 1} \right)\\
\Rightarrow 3\,\, \vdots \,\,\left( {2n + 1} \right)\\
n \in Z \Rightarrow \left( {2n + 1} \right) \in Z \Rightarrow \left( {2n + 1} \right) \in Ư\left( 3 \right) = \left\{ { \pm 1; \pm 3} \right\}\\
\Rightarrow 2n \in \left\{ { - 4; - 2;0;2} \right\}\\
\Rightarrow n \in \left\{ { - 2; - 1;0;1} \right\}
\end{array}\)
