a, $P=1-$$\frac{x}{x+2}=$ $\frac{x+2-x}{x+2}=$ $\frac{2}{x+2}$
Ta có: $|x-3|=1⇒$ \(\left[ \begin{array}{l}x=2\\x=4\end{array} \right.\)
Tại x=2, ta có: $P=$$\frac{2}{2+2}=$ $\frac{1}{2}$
Tại x=4, ta có: $P=$$\frac{2}{4+2}=$ $\frac{1}{3}$
b, $Q=$$\frac{x}{x^2-4}+$ $\frac{1}{x+2}+$ $\frac{2}{2-x}$ $=>Q=$$\frac{x+1(x-2)-2(x+2)}{(x-2)(x+2)}=>Q=$ $\frac{-6}{(x-2)$
$Ta có:A=Q:P$$(x+2)}$:$\frac{2}{x+2}=>A=$ $\frac{-6(x+2)}{2(x-2)(x+2)}=>A=$ $\frac{-6}{2(x-2)}=>A=$ $\frac{-6}{2x-4}$
Để A= nguyên ⇒2x-4∈Ư(6)={±1;±2;±3;±6}
2x-4=1⇒x=5/2
2x-4=-1⇒x=3/2
2x-4=2⇒x=3
2x-4=-2⇒x=1
2x-4=3⇒x=7/2
2x-4=-3⇒x=1/2
2x-4=6⇒x=5
2x-4=-6⇒x=-1
Vậy x∈{5/2;3/2;3;1;7/2;1/2;5;-1}
