a, ĐK : x≥0,x khác 1
P= $\frac{3}{√x +1}$ -$\frac{1}{√x - 1}$ -$\frac{√x-5}{x-1}$
= $\frac{3(√x-1)}{(√x-1)(√+1)}$ -$\frac{1(√+1)}{(√x-1)(√x+1)}$ -$\frac{√x-5}{(√x-1)(√x+1)}$
=$\frac{3√x-3-√x-1-√x+5}{(√x-1)(√x+1)}$
=$\frac{√x+1}{(√x-1)(√x+1)}$
=$\frac{1}{√x-1}$