a) A = $\frac{(x+1)(x+2)+6x-9x(x+1)}{3x(x+1)}$. $\frac{x+1}{2-4x}$ - $\frac{3x-x^2+1}{3x}$ (đk: x $\neq$ 0; x $\neq$ -1)
= $\frac{x^2+2x+x+2+6x-9x^2-9x}{3x}$.$\frac{1}{2(1-2x)}$ - $\frac{3x-x^2+1}{3x}$
= $\frac{-8x^2+2}{3x}$.$\frac{1}{2(1-2x)}$ - $\frac{3x-x^2+1}{3x}$
= $\frac{2(-4x^2+1)}{3x}$. $\frac{1}{2(1-2x)}$ - $\frac{3x-x^2+1}{3x}$
= $\frac{1-4x^2}{3x}$.$\frac{1}{1-2x}$ - $\frac{3x-x^2+1}{3x}$
= $\frac{(1-2x)(1+2x)}{3x}$.$\frac{1}{1-2x}$ - $\frac{3x-x^2+1}{3x}$
= $\frac{1+2x}{3x}$ $\frac{3x-x^2+1}{3x}$
= $\frac{1+2x-(3x-x^2+1)}{3x}$
= $\frac{1+2x-3x+x^2-1}{3x}$
= $\frac{-x+x^2}{3x}$
= $\frac{x(-1+x)}{3x}$
= $\frac{-1+x}{3}$
Vậy A = $\frac{-1+x}{3}$ với x $\neq$ 0; x $\neq$ -1
b) A = 3
↔ $\frac{-1+x}{3}$ = 3 (đk: x $\neq$ 0; x $\neq$ -1)
↔ -1 + x = 9
↔ x = 9 +1
↔ x = 10 (tm)
Vậy x = 10 khi A = $\frac{-1+x}{3}$ với x $\neq$ 0; x $\neq$ -1
c) /x/ = 1
↔\(\left[ \begin{array}{l}x=1(tm)\\x=-1(ktm)\end{array} \right.\)
Thay x = 1 vào biểu thức A ta có:
A = $\frac{-1+1}{3}$
= 0 : 3
= 0
Vậy A = 0 khi x = 1 với x $\neq$ 0; x $\neq$ -1
