$\text{a, Để P = 1}$
`⇒ \frac{2\sqrt{x}}{3x-1} = 1` `(x ≥ 0; x ne 1/3)`
`⇔ \frac{2\sqrt{x}}{3x-1} - 1 = 0`
`⇔ \frac{2\sqrt{x}-3x+1}{3x-1}= 0`
`⇒ 2\sqrt{x}-3x+1=0`
`⇔ 3\sqrt{x}-\sqrt{x}-3x+1=0`
`⇔ -3\sqrt{x}(\sqrt{x}-1)-(\sqrt{x}-1)=0`
`⇔ -(3\sqrt{x}+1)(\sqrt{x}-1)=0`
`⇔ (3\sqrt{x}+1)(\sqrt{x}-1)=0`
`⇔` \(\left[ \begin{array}{l}3\sqrt{x}+1=0\\\sqrt{x}-1=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x∈∅\\x=1(TM)\end{array} \right.\)
$\text{Vậy để P = 1 thì x = 1}$
$\text{b, Để P ≥ 0}$
`⇒ \frac{2\sqrt{x}}{3x-1} ≥ 0`
$\text{Có: $\sqrt{x}$ ≥ 0 ∀ x ⇒ 2$\sqrt{x}$ ≥ 0 ∀ x}$
`⇒ 3x - 1 ≥ 0 ⇔ x ≥ 1/3`
$\text{Vậy để P ≥ 0 thì x ≥ $\dfrac{1}{3}$}$
