Lời giải chi tiết:

Đặt \(f\left( x \right) = a{x^4} + b{x^3} + c{x^2} + dx + e\)

\( \Rightarrow \left\{ \begin{array}{l}f'\left( x \right) = 4a{x^3} + 3b{x^2} + 2cx + d\\f''\left( x \right) = 12a{x^2} + 6bx + 2c = 2\left( {6a{x^2} + 3bx + c} \right)\end{array} \right.\)

Theo bài ra ta có \({\left[ {f'\left( x \right)} \right]^2} - f''\left( x \right)f\left( x \right) = 0 \Leftrightarrow \dfrac{{{{\left[ {f'\left( x \right)} \right]}^2} - f''\left( x \right)f\left( x \right)}}{{{f^2}\left( x \right)}} = 0 \Leftrightarrow \left( {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}} \right)' = 0\).

Phương trình \(a{x^4} + b{x^3} + c{x^2} + dx + e = 0\) giả sử có 4 nghiệm phân biệt \({x_1},\,\,{x_2},\,\,{x_3},\,\,{x_4}\)

\( \Rightarrow f\left( x \right) = a\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\left( {x - {x_3}} \right)\left( {x - {x_4}} \right)\).

Ta có \(f'\left( x \right) = a\left[ \begin{array}{l}\left( {x - {x_2}} \right)\left( {x - {x_3}} \right)\left( {x - {x_4}} \right) + \left( {x - {x_1}} \right)\left( {x - {x_3}} \right)\left( {x - {x_4}} \right)\\\,\,\,\,\,\,\, + \left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\left( {x - {x_4}} \right) + \left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\left( {x - {x_3}} \right)\end{array} \right]\)

\(\begin{array}{l} \Rightarrow \dfrac{{f'\left( x \right)}}{{f\left( x \right)}} = \dfrac{1}{{x - {x_1}}} + \dfrac{1}{{x - {x_2}}} + \dfrac{1}{{x - {x_3}}} + \dfrac{1}{{x - {x_4}}}\\ \Rightarrow \left[ {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}} \right]' = \dfrac{{ - 1}}{{{{\left( {x - {x_1}} \right)}^2}}} + \dfrac{{ - 1}}{{{{\left( {x - {x_2}} \right)}^2}}} + \dfrac{{ - 1}}{{{{\left( {x - {x_3}} \right)}^2}}} + \dfrac{{ - 1}}{{{{\left( {x - {x_4}} \right)}^2}}} < 0\end{array}\)

Vậy phương trình \(\left( {\dfrac{{f'\left( x \right)}}{{f\left( x \right)}}} \right)' = 0\) vô nghiệm hay phương trình

\({\left( {4a{x^3} + 3b{x^2} + 2cx + d} \right)^2} - 2\left( {6a{x^2} + 3bx + c} \right)\left( {a{x^4} + b{x^3} + c{x^2} + dx + e} \right) = 0\)  vô nghiệm.

Chọn A.