Câu hỏi 3

Ta có \(f'\left( x \right) = {x^3}{f^2}\left( x \right)\,\, \Leftrightarrow \frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}} = {x^3}\)

Lấy nguyên hàm hai vế ta được \(\int {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}dx}  = \int {{x^3}dx}  \Leftrightarrow \int {\frac{1}{{{f^2}\left( x \right)}}d\left( {f\left( x \right)} \right)}  = \frac{{{x^4}}}{4} + C\)

\( \Leftrightarrow  - \frac{1}{{f\left( x \right)}} = \frac{{{x^4}}}{4} + C \Rightarrow f\left( x \right) = \frac{{ - 1}}{{\frac{{{x^4}}}{4} + C}}\)

Mà \(f\left( 2 \right) =  - \frac{4}{{19}}\) nên \(f\left( 2 \right) = \frac{{ - 1}}{{\frac{{{2^4}}}{4} + C}} \Leftrightarrow \frac{{19}}{4} = 4 + C \Leftrightarrow C = \frac{3}{4} \Leftrightarrow f\left( x \right) = \frac{{ - 4}}{{{x^4} + 3}}\)

Suy ra \(f\left( 1 \right) = \frac{{ - 4}}{{{1^4} + 3}} =  - 1.\)

Chọn C.