Đáp án:
$S = \{0;1\}$
Giải thích các bước giải:
$1+\dfrac{2}{6} \sqrt{x-x^2} = \sqrt{x} + \sqrt{1-x}$
ĐK : $0 \le x \le 1$
$\Leftrightarrow 1 + \dfrac{2}{3} \sqrt{x}. \sqrt{1-x} = \sqrt{x} +\sqrt{1-x}$
Đặt $\sqrt{x} + \sqrt{1-x} = t (t >0)$
$\to \sqrt{x}. \sqrt{1-x} = \dfrac{t^2-1}{2}$
$\to 1 + \dfrac{t^2-1}{2} = t$
$\to t^2 - 2t +1=0$
$\to t =1$
$\to \sqrt{x} + \sqrt{1-x} = 1$
$\to x = (1-\sqrt{1-x})^2$
$\to x = 1-2\sqrt{1-x}+1-2x$
$\to 2-2x-2\sqrt{1-x}=0$
$\to 4-8x+4x^2=4(1-x)$
$\to 4x^2-4x=0$
$\to \left[ \begin{array}{l}x=0\\x=1\end{array} \right.$
Vậy $S = \{0;1\}$
