Đề không rõ nên mk chia 2 trường hợp
TH1:
$P$=$\dfrac{1}{2}$ ⇔ $2\sqrt[]{x}$ = $\sqrt[]{x}+3$ ($x$\geq$ $0$)
⇔ $2\sqrt[]{x}$-$\sqrt[]{x}$=$3$
⇔ $\sqrt[]{x}$=$3$
⇔ $(\sqrt[]{x})^2$=$9$
⇔ $x=9$
TH2:
$P$=$\dfrac{1}{2}$ ⇔ $2\sqrt[]{x}$ = $\sqrt[]{x+3}$ ($x$\geq$ $0$)
⇔ $(2\sqrt[]{x})^2$ = $(\sqrt[]{x+3})^2$
⇔ $4x$=$x+3$
⇔ $4x-x$=$3$
⇔ $3x=3$
⇔ $x=1$