Gọi \({x_1},{\rm{ }}{x_2}{\rm{ }}\left( {{x_1} < {x_2}} \right)\) là hai nghiệm của phương trình \(\left| {{x^2} - 4x - 5} \right| = 4x - 17\). Tính giá trị biểu thức \(P = x_1^2 + {x_2}.\)

Phương trình \( \Leftrightarrow \left\{ \begin{array}{l}4x - 17 \ge 0\\{\left| {{x^2} - 4x - 5} \right|^2} = {\left( {4x - 17} \right)^2}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}x \ge \dfrac{{17}}{4}\\{\left( {{x^2} - 4x - 5} \right)^2} = {\left( {4x - 17} \right)^2}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x \ge \dfrac{{17}}{4}\\\left( {{x^2} - 8x + 12} \right)\left( {{x^2} - 22} \right) = 0\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}x \ge \dfrac{{17}}{4}\\\left[ {\begin{array}{*{20}{c}}{{x^2} - 8x + 12 = 0}\\{{x^2} - 22 = 0}\end{array}} \right.\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x \ge \dfrac{{17}}{4}\\\left[ {\begin{array}{*{20}{c}}{x = 2 \vee x = 6}\\{x =  \pm \sqrt {22} }\end{array}} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 6\\x = \sqrt {22} \end{array} \right.\)\( \Rightarrow P = {\left( {\sqrt {22} } \right)^2} + 6 = 28\)