`a, \frac{2\sqrt{x}}{\sqrt{x}+3}(1+\frac{1}{\sqrt{x}+2})+\frac{9\sqrt{x}+14}{x+3\sqrt{x}+2}` `(x ≥ 0)`
`⇔ \frac{2\sqrt{x}}{\sqrt{x}+3}.\frac{\sqrt{x}+2+1}{\sqrt{x}+2} + \frac{9\sqrt{x}+14}{x+\sqrt{x}+2\sqrt{x}+2}`
`⇔ \frac{2\sqrt{x}}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{\sqrt{x}+2}+\frac{9\sqrt{x}+14}{\sqrt{x}(\sqrt{x}+1)+2(\sqrt{x}+1)}`
`⇔ \frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{9\sqrt{x}+14}{(\sqrt{x}+1)(\sqrt{x}+2)`
`⇔ \frac{2\sqrt{x}(\sqrt{x}+1)+9\sqrt{x}+14}{(\sqrt{x}+2)(\sqrt{x}+1)}`
`⇔ \frac{2x+2\sqrt{x}+9\sqrt{x}+14}{(\sqrt{x}+2)(\sqrt{x}+1)}`
`⇔ \frac{2x+11\sqrt{x}+14}{(\sqrt{x}+2)(\sqrt{x}+1)}`
`⇔ \frac{2x+4\sqrt{x}+7\sqrt{x}+14}{(\sqrt{x}+2)(\sqrt{x}+1)}`
`⇔ \frac{2\sqrt{x}(\sqrt{x}+2)+7(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}+1)}`
`⇔ \frac{(2\sqrt{x}+7)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}+1)`
`⇔ \frac{2\sqrt{x}+7}{\sqrt{x}+1}`
b, $\text{Thay x = 4 (TM) vào P}$
`⇒ P = \frac{2.\sqrt{4}+7}{\sqrt{4}+1} = \frac{2.2+7}{2+1}=11/3`
$\text{Vậy tại x = 4 thì P = $\dfrac{11}{3}$}$
$\text{c, Có: P = $\dfrac{2\sqrt{x}+7}{\sqrt{x}+1}$ = 2 + $\dfrac{5}{\sqrt{x}+1}$}$
$\text{Để P ∈ N}$
`⇒ \sqrt{x}+1 ∈ Ư(5)={±1; ±5}`
$\text{Mà $\sqrt{x}$+1 > 0 ⇒ $\sqrt{x}$ + 1 ∈ {1; 5}}$
`⇒ x ∈ {0; 2}`
