a) $A$ có nghĩa khi:
$\left\{ \begin{array}{l}\sqrt[]{x}≥0\\\sqrt[]{x}+1 \neq 0\end{array} \right.$
$→ x≥0$
$B$ có nghĩa khi:
$\left\{ \begin{array}{l}\sqrt[]{x}≥0\\x-1 \neq 0\end{array} \right.$
$↔ \left\{ \begin{array}{l}x≥0\\x \neq 1\end{array} \right.$
b) Thay $x=9$ vào $A$, ta có:
$A=\dfrac{\sqrt[]{9}-1}{\sqrt[]{9}+1}=\dfrac{3-1}{3+1}=\dfrac{1}{2}$
c) $B=\dfrac{\sqrt[]{x}+3}{\sqrt[]{x}+1}-\dfrac{4}{1-\sqrt[]{x}}+\dfrac{5-x}{x-1}$
$=\dfrac{(\sqrt[]{x}+3)(\sqrt[]{x}-1)+4(\sqrt[]{x}+1)+5-x}{(\sqrt[]{x}+1)(\sqrt[]{x}-1)}$
$=\dfrac{x+2\sqrt[]{x}-3+4\sqrt[]{x}+4+5-x}{(\sqrt[]{x}+1)(\sqrt[]{x}-1)}$
$=\dfrac{6(\sqrt[]{x}+1)}{(\sqrt[]{x}+1)(\sqrt[]{x}-1)}$
$=\dfrac{6}{\sqrt[]{x}-1}$
$→ P=A.B$
$=\dfrac{\sqrt[]{x}-1}{\sqrt[]{x}+1}.\dfrac{6}{\sqrt[]{x}-1}$
$=\dfrac{6}{\sqrt[]{x}+1}$
