Giải thích các bước giải:
a) $\left ( \dfrac{x^{2} - 3x}{x^{2} - 9} - 1 \right ) : \left ( \dfrac{9 - x^{2}}{x^{2} + x - 6} - \dfrac{x - 3}{2 - x} - \dfrac{x - 2}{x + 3} \right )$ (ĐK $x \neq 2; x \neq 3; x \neq -3$)
$= \left [ \dfrac{x\left ( x - 3 \right )}{\left ( x + 3 \right )\left ( x - 3 \right )} \right ] : \left [ \dfrac{9 - x^{2}}{\left ( x - 2 \right )\left ( x + 3 \right )} + \dfrac{x - 3}{x - 2} - \dfrac{x - 2}{x + 3} \right ]$
$= \left ( \dfrac{x}{x + 3} - 1 \right ) : \dfrac{9 - x^{2} + \left ( x - 3 \right )\left ( x + 3 \right ) - \left ( x - 2 \right )^{2}}{\left ( x - 2 \right )\left ( x + 3 \right )}$
$= \dfrac{x - x - 3}{x + 3} : \dfrac{9 - x^{2} + x^{2} - 9 - x^{2} + 4x - 4}{\left ( x - 2 \right )\left ( x + 3 \right )}$
$= \dfrac{3}{x + 3} : \dfrac{x^{2} - 4x + 4}{\left ( x - 2 \right )\left ( x + 3 \right )}$
$= \dfrac{3}{x + 3}.\dfrac{\left ( x - 2 \right )\left ( x + 3 \right )}{x^{2} - 4x + 4}$
$= \dfrac{3}{x + 3}.\dfrac{\left ( x - 2 \right )\left ( x + 3 \right )}{\left ( x - 2 \right )^{2}}$
$= \dfrac{3}{x - 2}$
b) $x^{2} - 3x + 2 = 0$
$\Leftrightarrow x^{2} - 2x - x + 2 = 0$
$\Leftrightarrow x\left ( x - 2 \right ) - \left ( x - 2 \right ) = 0$
$\Leftrightarrow \left ( x - 2 \right )\left ( x - 1 \right ) = 0$
$\Leftrightarrow \left[ \begin{array}{l}x - 2 = 0\\x - 1 = 0\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x = 2 (L)\\x = 1 (tm)\end{array} \right.$
Thay $x = 1$ ta có:
$\dfrac{3}{x - 2} = \dfrac{3}{1 - 2} = -3$
