Phương pháp giải:

 \(\begin{array}{l}y = f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{f_1}\left( x \right){\rm{ }}\,\,khi\,\,\,\,\,x \in {D_1}{\rm{ }}}\\{{f_2}\left( x \right)\,\,\,{\rm{ }}khi\,\,\,x \in {D_2}}\\{{f_3}\left( x \right){\rm{ }}\,khi\,\,\,x \in {D_3}}\end{array}} \right.\\{D_f} = {D_1} \cup {D_2} \cup {D_3}\\f\left( {{x_1}} \right) = {f_1}\left( {{x_1}} \right){\rm{ }};{\rm{ }}{x_1} \in {D_1}\\f\left( {{x_2}} \right) = {f_2}\left( {{x_2}} \right){\rm{ ; }}{x_2} \in {D_2}\\f\left( {{x_3}} \right) = {f_3}\left( {{x_3}} \right){\rm{ ; }}{x_3} \in {D_3}\end{array}\)

\({x_4} \notin D \Rightarrow \) không tồn tại \(f\left( {{x_4}} \right).\)

Lời giải chi tiết:

Ta có: \(\left\{ \begin{array}{l}f\left( 2 \right) = \frac{{2\sqrt {2 + 2}  - 3}}{{2 - 1}} = 1\\f\left( { - 2} \right) = {\left( { - 2} \right)^2} + 1 = 5\end{array} \right. \Rightarrow P = 1 + 5 = 6.\)  

Chọn  C.