Câu hỏi 37

\(CF = \sqrt {{x^2} - {{\left( {8 - x} \right)}^2}}  = \sqrt {16x - 64} \)

đồng dạng với \(\Delta FCE \Rightarrow \dfrac{{EF}}{{AF}} = \dfrac{{CF}}{{AD}} \Rightarrow AF = \dfrac{{EF.AD}}{{CF}} = \dfrac{{8x}}{{\sqrt {16x - 64} }}\)

\(\begin{array}{l}AE = y = \sqrt {A{F^2} + E{F^2}}  = \sqrt {\dfrac{{64{x^2}}}{{16x - 64}} + {x^2}}  = \sqrt {\dfrac{{16{x^3}}}{{16x - 64}}} \\AE\,\min  \Leftrightarrow {y^2}\min \\{y^2} = f\left( x \right) = \dfrac{{16{x^3}}}{{16x - 64}} \Rightarrow f'\left( x \right) = \dfrac{{48{x^2}\left( {16x - 64} \right) - 16.16{x^3}}}{{{{\left( {16x - 64} \right)}^2}}}\\ \Leftrightarrow x = 6\end{array}\)

BBT:

\({y_{\min }} = \sqrt {f{{\left( x \right)}_{\min }}}  = \sqrt {108}  = 6\sqrt 3 \)

Chọn A.