a)
ĐKXĐ của A: x khác -2
A= ($\dfrac{4x}{x+2}$ - $\dfrac{x^3-8}{x^3+8}$.$\dfrac{4x^2-8x+16}{x^2-4}$) : $\dfrac{16}{x+2}$ . $\dfrac{x^2+x+2}{x^2+x+1}$
= [$\dfrac{4x}{x+2}$ - $\dfrac{(x-2)(x^2+2x+4)}{(x+2)(x^2-2x+4)}$ . $\dfrac{4(x^2-2x+4)}{(x-2)(x+2)}$ . $\dfrac{x+2}{16}$ . $\dfrac{(x+1)(x+2)}{x^2+x+1}$
= [ $\dfrac{4x}{x+2}$ - $\dfrac{4(x^2+2x+4)}{(x+2)^2}$] . $\dfrac{x+2}{16}$ . $\dfrac{(x+1)(x+2)}{x^2+x+1}$
= $\dfrac{4x(x+2)-4(x^2+2x+4)}{(x+2)^2}$ . $\dfrac{x+2}{16}$ . $\dfrac{(x+1)(x+2)}{x^2+x+1}$
= $\dfrac{-x-1}{x^2+x+1}$
ĐKXĐ của B: x khác 1
B= $\dfrac{x^2+x-2}{x^3-1}$
= $\dfrac{(x-1)(x+2)}{(x-1)(x^2+x+1)}$
= $\dfrac{x+2}{x^2+x+1}$
b)
A+B= $\dfrac{-x-1+x+2}{x^2+x+1}$
= $\dfrac{1}{x^2+x+1}$
= $\dfrac{1}{(x+1/2)^2+3/4}$ \leq 1/(3/4) =4/3
Vậy, max(A+B) = 4/3 \Leftrightarrow x= -1/2
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
