Đáp án: a) P= ($\frac{4}{x-4}$ - $\frac{4}{x+4}$) : $\frac{32}{x²+8x+16}$ (x$\neq$ 4)
= ($\frac{4.(x+4)}{(x-4)}$ - $\frac{4.(x-4)}{(x+4)(x-4)}$) : $\frac{32}{x²+2.x.4+4²}$
= $\frac{4.(x+4)-4.(x-4)}{(x+4)(x-4)}$ . $\frac{(x+4)²}{32}$
= $\frac{4.(x+4-x+4)}{x-4}$ . $\frac{x+4}{32}$
= $\frac{4.8}{x-4}$ . $\frac{x+4}{32}$
= $\frac{x+4}{x-4}$
b) để P nguyên hay P ∈ Z ⇔$\frac{x+4}{x-4}$ ∈ Z
Ta có: $\frac{x+4}{x-4}$ = $\frac{x-4+8}{x-4}$
= $\frac{x-4}{x-4}$ + $\frac{8}{x-4}$
= 1+$\frac{8}{x-4}$
⇒ để P ∈ Z thì x-4 ∈ Ư(8)={1;-1;2;-2;4;-4;8;-8}
=> x-4 = 1 ⇔ x = 5
x-4 = -1 ⇔ x= 3
x-4 = 2 ⇔ x= 6
x-4 = -2 ⇔ x= 2
x-4 = 4 ⇔ x= 8
x-4 = -4 ⇔ x= 0
x-4 = 8 ⇔ x= 12
x-4 = -8 ⇔ x= -4
