A = $\frac{x-2}{x^3-x^2-x-2}$ (Đk: x $\neq$ 2)
= $\frac{x-2}{(x^3-8)-(x^2-4x+4)-5(x-2)}$
= $\frac{x-2}{(x-2)(x^2+2x+4)-(x-2)^2-5(x-2)}$
= $\frac{x-2}{(x-2)(x^2+2x+4-x+2-5)}$
= $\frac{x-2}{(x-2)(x^2+x+1)}$
= $\frac{1}{x^2+x+1}$
⇒ $\frac{1}{A}$ = $x^{2}$ + x + 1 = $(x+\frac{1}{2})^{2}$ + $\frac{3}{4}$ $\geq$ $\frac{3}{4}$
⇒ A $\leq$ $\frac{4}{3}$
Dấu "=" xảy ra khi $(x+\frac{1}{2})^{2}$ = 0 ⇒ x = $\frac{-1}{2}$
Vậy $A_{Max}$ = $\frac{4}{3}$ khi x = $\frac{-1}{2}$
