$A$ = ($\frac{1}{√x-1}$ + $\frac{√x}{1-x}$).$\frac{x-√x}{2√x+1}$ (x ≥ 0; x $\neq$ 1)
= ($\frac{1}{√x-1}$ - $\frac{√x}{x-1}$).$\frac{√x(√x-1)}{2√x+1}$
= ($\frac{√x + 1 - √x}{x-1}$).$\frac{√x(√x-1)}{2√x+1}$
= $\frac{√x(√x-1)}{(x - 1).(2√x+1)}$
= $\frac{√x}{(√x + 1).(2√x+1)}$
= $\frac{√x}{2x+√x+2√x+2}$
= $\frac{√x}{2x+3√x+2}$
Vậy $A$ = $\frac{√x}{2x+3√x+2}$
