Cho hàm số \(f\left( x \right)\) thỏa \(f\left( 1 \right) = \dfrac{1}{3}\)và \(f'\left( x \right) = {\left[ {xf\left( x \right)} \right]^2}\) với mọi \(x \in \mathbb{R}\). Giá trị \(f\left( 2 \right)\) bằng

Ta có \(f'\left( x \right) = {\left[ {xf\left( x \right)} \right]^2}\)\( \Leftrightarrow \dfrac{{f'\left( x \right)}}{{{f^2}\left( x \right)}} = {x^2}\)

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

Mà \(f\left( 1 \right) = \dfrac{1}{3} \Rightarrow  - \dfrac{1}{{\dfrac{1}{3}}} = \dfrac{1}{3} + C \Leftrightarrow C = \dfrac{{ - 10}}{3}\)

\(\begin{array}{l} \Rightarrow  - \dfrac{1}{{f\left( x \right)}} = \dfrac{{{x^3}}}{3} - \dfrac{{10}}{3} = \dfrac{{{x^3} - 10}}{3}\\ \Rightarrow f\left( x \right) = \dfrac{3}{{10 - {x^3}}}\end{array}\)

Vậy \(f\left( 2 \right) = \dfrac{3}{{10 - {2^3}}} = \dfrac{3}{2}.\)