Đáp án:
ĐKXĐ: x>0, x$\neq$ 9
A= √x ($\frac{√x}{3+√x}$ +$\frac{x+9}{9-x}$). ($\frac{3√x+1}{x-3√x}$ -$\frac{1}{√x}$ )
=($\frac{√x(3-√x)}{(3+√x)(3-√x)}$ +$\frac{x+9}{9-x}$). ($\frac{3√x+1}{x-3√x}$ -$\frac{√x-3}{√x(√x-3)}$ )
=($\frac{3√x-x}{9-x}$ +$\frac{x+9}{9-x}$). ($\frac{3√x+1}{x-3√x}$ -$\frac{√x-3}{x-3√x}$ )
=$\frac{3√x+9}{9-x}$ .$\frac{2√x+4}{x-3√x}$
=$\frac{3(√x+3)}{(√x+3)(√x-3)}$ .$\frac{2√x+4}{x-3√x}$
=$\frac{6√x+12}{√x(√x-3)^{2}}$
