Đáp án đúng là: A

Từ giả thiết, ta có: \[{\left( {{x^2} + 1} \right)^2}f'\left( x \right) = {\left[ {f\left( x \right)} \right]^2}\left( {{x^2} - 1} \right)\]

\[ \Leftrightarrow f'\left( x \right) = \frac{{{{\left[ {f\left( x \right)} \right]}^2}\left( {{x^2} - 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} > 0,\forall x \in \left[ {1;2} \right]\].

Xét với mọi \[x \in \left[ {1;2} \right]\], ta có:

\[\frac{{f'\left( x \right)}}{{{{\left[ {f\left( x \right)} \right]}^2}}} = \frac{{\left( {{x^2} - 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}} \Leftrightarrow \int {\frac{{f'\left( x \right)}}{{{{\left[ {f\left( x \right)} \right]}^2}}}dx} = \int {\frac{{\left( {{x^2} - 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx\].

\[ \Rightarrow \int {\frac{{f'\left( x \right)}}{{{{\left[ {f\left( x \right)} \right]}^2}}}dx} = \int {\frac{{\left( {1 - \frac{1}{{{x^2}}}} \right)}}{{{{\left( {x + \frac{1}{x}} \right)}^2}}}} dx = \int {\frac{{d\left( {x + \frac{1}{x}} \right)}}{{{{\left( {x + \frac{1}{x}} \right)}^2}}}} \]

\[ \Rightarrow - \frac{1}{{f\left( x \right)}} = \frac{{ - 1}}{{x + \frac{1}{x}}} + C \Leftrightarrow f\left( x \right) = x + \frac{1}{x} + C.\]

Mà \[f\left( 1 \right) = 1 \Leftrightarrow 1 = 2 + C \Leftrightarrow C = - 1\].

Vậy \[f\left( x \right) = x + \frac{1}{x} - 1\].

Suy ra \[f\left( 2 \right) = 2 + \frac{1}{2} - 1 = \frac{3}{2}.\]