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