a) ĐKXĐ:
+) x² - 9 $\neq$ 0
⇔ x² $\neq$ 9
⇔ x $\neq$ ±3
+) x² + 1 $\neq$ 0
⇔ x² $\neq$ -1
⇔ Có vô số nghiệm x thỏa mãn (Vì x² ≥ 0 ∀ x ∈ Z)
b) $A$ = ($\frac{2x²}{x²-9}$ - $\frac{x}{x-3}$).$\frac{x+3}{x²+1}$
= ($\frac{2x²}{(x - 3)(x+3)}$ - $\frac{x(x + 3)}{(x-3)(x+3)}$).$\frac{x+3}{x²+1}$
= $\frac{2x²-x(x+3)}{(x-3)(x+3)}$.$\frac{x+3}{x²+1}$
= $\frac{2x²-x²-3x}{(x-3)(x+3)}$.$\frac{x+3}{x²+1}$
= $\frac{x²-3x}{(x-3)(x+3)}$.$\frac{x+3}{x²+1}$
= $\frac{x(x-3)}{(x-3)(x+3)}$.$\frac{x+3}{x²+1}$
= $\frac{x}{(x+3)}$.$\frac{x+3}{x²+1}$
= $\frac{x.(x+3)}{(x+3).(x²+1)}$
= $\frac{x}{x²+1}$
Vậy $A$ = $\frac{x}{x²+1}$
c) Để $A$ = $\frac{x}{x²+1}$ = $1/2$
⇔ $\frac{x}{x²+1}$ = $1/2$
`⇔ 2x = x² + 1`
`⇔ x² + 1 - 2x = 0`
`⇔ (x - 1)² = 0`
`⇔ x - 1 = 0`
`⇔ x = 1`
Vậy x = 1 thì $A$ = $1/2$
