P=(1+$\frac{√x}{√x+1}$ ):($\frac{1}{√x-1}$ - $\frac{2√x}{x√x+√x-1}$ )-$\frac{x√x+1}{x-√x+1}$
= $\frac{√x+1+√x}{√x+1}$ : ($\frac{1}{√x-1}$ - $\frac{2√x}{√x(x+1)-(x+1)}$ ) - $\frac{x√x+1}{x-√x+1}$
= $\frac{2√x}{√x+1}$ :($\frac{1}{√x-1}$- $\frac{2√x}{(x+1)(√x-1)}$ )-$\frac{x√x+1}{x-√x-1}$
=$\frac{2√x}{√x+1}$ :($\frac{x-2√x+1}{(x+1)(√x-1)}$ )-$\frac{x√x+1}{x-√x-1}$
= $\frac{2√x}{√x+1}$ :($\frac{(√x-1)^2}{(x+1)(√x-1)}$ )-$\frac{x√x+1}{x-√x+1}$
=$\frac{2√x}{√x+1}$ . $\frac{x+1}{√x-1}$ -$\frac{x√x+1}{x-√x-1}$
=$\frac{2x√x+2√x}{x-1}$ - $\frac{x√x+1}{x-√x-1}$
=$\frac{x-√x-1}{(x-1)(x-√x-1)}$
=$\frac{1}{x-1}$
Đẻ P <4 thì
$\frac{1}{x-1}$ <4
<=> $\frac{1}{x-1}$ -4<0
<=> x>$\frac{6}{5}$
Để P đạt giá trị nguyên thì frac{1}{x-1}$ nguyên
=> x-1 ∈ (1)
+,x-1=-1 => x=0 (tm)
+,x-1=1=>x=2 (tm)
Vậy x=0 hoặc x=2 để P đạt giá trị nguyên
