a) $x(x-3)=0$
$↔ \left[ \begin{array}{l}x=0\\x-3=0\end{array} \right.$
$↔ \left[ \begin{array}{l}x=0\\x=3\end{array} \right.$
Thay $x=0$ vào $A$, ta có:
$A=\dfrac{0}{0-2}=0$
Thay $x=3$ vào $A$, ta có:
$A=\dfrac{3}{3-2}=3$
b) $B=\dfrac{x}{x-2}-\dfrac{4}{x^2-2x}-\dfrac{4}{x}$
$=\dfrac{x^2-4-4(x-2)}{x(x-2)}$
$=\dfrac{x^2-4x+4}{x(x-2)}$
$=\dfrac{(x-2)^2}{x(x-2)}$
$=\dfrac{x-2}{x}$
c) $P=A:B$
$↔ P=\dfrac{x}{x-2}.\dfrac{x}{x-2}$
$=(\dfrac{x}{x-2})^2$
Vì $P≥0$ nên $P≤0$ khi và chỉ khi $P=0$
$→ \dfrac{x}{x-2}=0 ↔ x=0$.
