\(\begin{array}{l}

P = \dfrac{\sqrt x +1}{\sqrt x  - 2} + \dfrac{2\sqrt x }{\sqrt x + 2} + \dfrac{2 + 5\sqrt x}{4 - x}\\
a)\\
ĐKXĐ:\begin{cases}x\geqslant 0\\\sqrt x - 2 \ne 0\\\sqrt x + 2 \ne 0\\4 - x \ne 0\end{cases} \Leftrightarrow \begin{cases}x\geqslant 0\\x \ne 4\end{cases}\\
b)\\
P = \dfrac{\left(\sqrt x+ 1\right)\left(\sqrt x  +2\right)}{\left(\sqrt x - 2\right)\left(\sqrt x+ 2\right)} + \dfrac{2\sqrt x\left(\sqrt x  -2\right)}{\left(\sqrt x - 2\right)\left(\sqrt x+ 2\right)} - \dfrac{2 + 5\sqrt x}{\left(\sqrt x - 2\right)\left(\sqrt x+ 2\right)}\\
\to P = \dfrac{x + 3\sqrt x + 2 + 2x - 4\sqrt x -2 - 5\sqrt x}{\left(\sqrt x - 2\right)\left(\sqrt x+ 2\right)}\\
\to P = \dfrac{3x - 6\sqrt x}{\left(\sqrt x - 2\right)\left(\sqrt x+ 2\right)}\\
\to P = \dfrac{3\sqrt x\left(\sqrt x - 2\right)}{\left(\sqrt x - 2\right)\left(\sqrt x+ 2\right)}\\
\to P = \dfrac{3\sqrt x}{\sqrt x + 2}\\
c)\\
\quad P = 2\\
\to \dfrac{3\sqrt x}{\sqrt x + 2} = 2\\
\to 3\sqrt x = 2\left(\sqrt x + 2\right)\\
\to \sqrt x =4\\
\to x = 16\quad \text{(nhận)}\\
\text{Vậy}\ x = 16
\end{array}\)

`a)ĐKXĐ:{(x>=0),(4-x ne 0):}<=>{(x>=0),(x ne 4):}`

`b)P=(\sqrt[x]+1)/(\sqrt[x]-2)+(2\sqrt[x])/(\sqrt[x]+2)+(2+5\sqrt[x])/(4-x)`

`=(\sqrt[x]+1)/(\sqrt[x]-2)+(2\sqrt[x])/(\sqrt[x]+2)-(2+5\sqrt[x])/(x-4)`

`=((\sqrt[x]+1)(\sqrt[x]+2)+2\sqrt[x](\sqrt[x]-2)-2-5\sqrt[x])/((\sqrt[x]-2)(\sqrt[x]+2))`

`=(x+2\sqrt[x]+\sqrt[x]+2+2x-4\sqrt[x]-2-5\sqrt[x])/((\sqrt[x]-2)(\sqrt[x]+2))`

`=(3x-6\sqrt[x])/((\sqrt[x]-2)(\sqrt[x]+2))`

`=(3\sqrt[x](\sqrt[x]-2))/((\sqrt[x]-2)(\sqrt[x]+2))`

`=(3\sqrt[x])/(\sqrt[x]+2)`

`c)P=2<=>(3\sqrt[x])/(\sqrt[x]+2)=2(x>=0;xne4)`

`<=>3\sqrt[x]=2(\sqrt[x]+2)`

`<=>3\sqrt[x]=2\sqrt[x]+4`

`<=>3\sqrt[x]-2\sqrt[x]=4`

`<=>\sqrt[x]=4`

`<=>x=16`

Vậy `P=2` khi `x=16`