$\begin{array}{l}1)\\ P={\displaystyle \frac{\sqrt{x}}{\sqrt{x}-1}}+{\displaystyle \frac{3}{\sqrt{x}+1}}-{\displaystyle \frac{6\sqrt{x}-4}{x-1}}(x\ge 0;x\ne 1)\\ P={\displaystyle \frac{\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}-1)-6\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+1)}}\\ P={\displaystyle \frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+1)}}\\ P={\displaystyle \frac{{(\sqrt{x}-1)}^2}{(\sqrt{x}-1)\left(\sqrt{x+1}\right)}}\\ P={\displaystyle \frac{\sqrt{x}-1}{\sqrt{x}+1}}\\ 2)\\ x=4\to \sqrt{x}=2\\ \to P={\displaystyle \frac{2-1}{2+1}}={\displaystyle \frac{1}{3}}\\ 3)\\ P=-1\\ \iff {\displaystyle \frac{\sqrt{x}-1}{\sqrt{x}+1}}=-1\\ \iff \sqrt{x}-1=-\sqrt{x}-1\\ \iff 2\sqrt{x}=0\\ \iff x=0\left(n\right)\\ 4)\\ P={\displaystyle \frac{\sqrt{x}+1-2}{\sqrt{x}+1}}=1-{\displaystyle \frac{2}{\sqrt{x}+1}}\\ \text{Có }\sqrt{x}+1\ge 1>0,\mathrm{\forall }x\ge 0;x\ne 1\\ \to -{\displaystyle \frac{2}{\sqrt{x}+1}}>0\\ \to 1-{\displaystyle \frac{2}{\sqrt{x}+1}}>1\\ \text{Vậy }P>1,\mathrm{\forall }x\ge 0;x\ne 1\end{array}$

1, P = (sqrtx)/(sqrtx - 1) + 3/(sqrtx +1) - (6sqrtx - 4)/(x -1) (x >=0 ; x\ne 1)

= (sqrtx (sqrtx +1))/((sqrtx -1)(sqrtx +1)) + (3(sqrtx -1))/((sqrtx +1)(sqrtx -1)) - (6sqrtx - 4)/((sqrtx -1)(sqrtx +1))

= (x + sqrtx + 3sqrtx -3 - 6sqrtx + 4)/((sqrtx -1)(sqrtx +1))

= (x -2sqrtx + 1)/((sqrtx -1)(sqrtx +1))

= ((sqrtx -1)^2)/((sqrtx -1)(sqrtx +1))

= (sqrtx -1)/(sqrtx +1)

Vậy với x >=0 ; x\ne 1 thì P = (sqrtx -1)/(sqrtx +1)

2, x = 4 (TM)

Thay vào P ta có :

P = (sqrtx -1)/(sqrtx +1)

= (sqrt4 -1)/(sqrt4 + 1)

= (2 -1 )/(2+1)

= 1/3

Vậy P = 1/3 khi x = 4

3, P = -1

<=> (sqrtx - 1)/(sqrtx +1) = -1

=> sqrtx -1 = -(sqrtx +1)

<=> 2sqrtx = 0

<=> sqrtx =0

<=> x= 0 (TM)
Vậy x= 0 thì P= -1

4, P= (sqrtx -1)/(sqrtx +1)

= (sqrtx + 1 - 2)/(sqrtx +1)

= 1-  2/(sqrtx +1)

sqrtx + 1 >= 0 + 1 > 0

=> 2/(sqrtx +1) > 0

=> P =1 - 2/(sqrtx +1) < 1
Vậy P < 1 khi x >=0 ; x\ne1