$ĐK: 0 \leq x \leq 1$
Đặt $a=\sqrt x,b=\sqrt{1-x}$
$\Rightarrow \left\{ \begin{array}{l} a^2+b^2=1(1)\\ 1+\frac{2}3ab=a+b(2) \end{array} \right.$
Bình phương hai vế, ta có
$(2) \Leftrightarrow 1+\frac{4}3ab+\frac{4}9a^2b^2=a^2+b^2+2ab$
$\Leftrightarrow \frac{4}9a^2b^2+-\frac{2}3ab=0$
$\Rightarrow ab=0 or ab=\frac{3}2$
$\Rightarrow \left\{ \begin{array}{l} ab=0\\ a+b=1 \end{array} \right. \Rightarrow a,b \Rightarrow x$
Hoặc $\left\{ \begin{array}{l} ab=\frac{3}2\\ a+b=2 \end{array} \right.$ (loại vì $PT X^2-2X+\frac{3}2=0$ vô nghiệm)