Giải PT vô tỷ: \ 4 + \sqrt {4 + \sqrt x } = x.
Giải:
Đặt \ \left\{ \begin{array}{l} a = \sqrt x \\ b = \sqrt {4 + \sqrt x } \\ a,b \ge 2 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {a^2} = 4 + b\\ {b^2} = 4 + a \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \left( {a - b} \right)\left( {a + b} \right) - \left( {a - b} \right) = 0\\ {b^2} = 4 + a \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \left( {a - b} \right)\left( {a + b - 1} \right) = 0\\ {b^2} = 4 + a \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} a = b\\ {a^2} - a - 4 = 0\\ a \ge 4 \end{array} \right. \Leftrightarrow \left( {a;b} \right) = \left( {\frac{{1 + \sqrt {17} }}{2};\frac{{1 + \sqrt {17} }}{2}} \right) \Leftrightarrow x = {a^2} = \frac{{9 + \sqrt {17} }}{2}\\ \left\{ \begin{array}{l} b = 1 - a\\ {a^2} - 3a - 3 = 0\\ a \ge 4 \end{array} \right. \Leftrightarrow \left( {a;b} \right) = \left( {\frac{{3 + \sqrt {21} }}{2}; - \frac{{1 + \sqrt {21} }}{2}} \right)\left( L \right) \end{array} \right. \Rightarrow S = \left\{ {\frac{{9 + \sqrt {17} }}{2}} \right\}
