Монгол Бодлогын Сан

Заавар: Квадрат язгуурын доорхи илэрхийлэл сөрөг биш тоо байна.

$a,b>0$ бол $a>b\Leftrightarrow a^2>b^2$ байна.

Бодолт: $$D\colon\left\{\begin{array}{l} 2-\sqrt{1-x}\ge 0\\ 1-x\ge 0\\ 2-x\ge 0 \end{array}\right.\Leftrightarrow\left\{\begin{array}{l} 2\ge\sqrt{1-x}\\ 1\ge x\\ 2\ge x \end{array}\right.\Leftrightarrow\left\{\begin{array}{l} 4\ge 1-x\\ 1\ge x\\ 2\ge x \end{array}\right.\Leftrightarrow\left\{\begin{array}{l} x\ge-3\\ 1\ge x \end{array}\right.$$ тул $-3\le x\le 1$.

$$\sqrt{2-\sqrt{1-x}}-\sqrt{2-x}>0\Leftrightarrow \sqrt{2-\sqrt{1-x}}>\sqrt{2-x}\Leftrightarrow 2-\sqrt{1-x}>2-x$$ $$\Leftrightarrow x>\sqrt{1-x}\Leftrightarrow\left\{\begin{array}{l} 1-x\ge 0\\ x>0\\x^2>1-x \end{array}\right.\Leftrightarrow\left\{\begin{array}{l} 1\ge x\\ x>0\\x^2+x-1>0 \end{array}\right.$$ $$x^2+x-1>0\Leftrightarrow \bigg(x-\dfrac{-1-\sqrt5}{2}\bigg)\bigg(x-\dfrac{-1+\sqrt5}{2}\bigg)>0$$ тул $\dfrac{-1+\sqrt5}{2}< x\le 1$ байна.