Đề bài: (Bài của bạn Tuhn Hnim hỏi)
Giải bất phương trình: $\ \frac{{x - \sqrt x }}{{1 - \sqrt {2\left( {{x^2} - x + 1} \right)} }} \ge 1.$
Giải:
Ta thấy: $\ \sqrt {2\left( {{x^2} - x + 1} \right)} = = \sqrt {2{{\left( {x - \frac{1}{2}} \right)}^2} + \frac{3}{2}} > 1 \Rightarrow 1 - \sqrt {2\left( {{x^2} - x + 1} \right)} < 0.$
Khi đó: $\ \frac{{x - \sqrt x }}{{1 - \sqrt {2\left( {{x^2} - x + 1} \right)} }} \ge 1 \Leftrightarrow x - \sqrt x \le 1 - \sqrt {2\left( {{x^2} - x + 1} \right)} \Leftrightarrow \sqrt x - x + 1 \ge \sqrt {2\left( {{x^2} - x + 1} \right)} .$\[ \Leftrightarrow \left\{ \begin{array}{l}
{\left( {\sqrt x - x + 1} \right)^2} \ge 2\left( {{x^2} - x + 1} \right)\\
\sqrt x - x + 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} - x + 2x\sqrt x - 2\sqrt x + 1 \le 0\\
\frac{{1 - \sqrt 5 }}{2} \le x \le \frac{{1 + \sqrt 5 }}{2}
\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}
\left( {{x^2} + 2x\sqrt x + x} \right) - 2\left( {x + \sqrt x } \right) + 1 \le 0\\
\frac{{1 - \sqrt 5 }}{2} \le x \le \frac{{1 + \sqrt 5 }}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + \sqrt x } \right)^2} - 2\left( {x + \sqrt x } \right) + 1 \le 0\\
\frac{{1 - \sqrt 5 }}{2} \le x \le \frac{{1 + \sqrt 5 }}{2}
\end{array} \right.\]\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{{{\left( {x + \sqrt x - 1} \right)}^2} \le 0}\\
{\frac{{1 - \sqrt 5 }}{2} \le x \le \frac{{1 + \sqrt 5 }}{2}}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x + \sqrt x - 1 = 0}\\
{\frac{{1 - \sqrt 5 }}{2} \le x \le \frac{{1 + \sqrt 5 }}{2}}
\end{array}} \right. \Leftrightarrow \sqrt x = \frac{{\sqrt 5 - 1}}{2} \Leftrightarrow S = \left\{ {\frac{{3 - \sqrt 5 }}{2}} \right\}\]
