Cho hàm số \(f(x)\) dương thỏa mãn \(f(0)=e\) và \({x^2}f\left( x \right) = f\left( x \right) + f\left( x \right),\forall x \ne&nbs

* Hướng dẫn giải

Ta có: \({x^2}f'\left( x \right) = f\left( x \right) + f'\left( x \right),\forall x \ne  \pm 1 \Rightarrow f'\left( x \right).\left( {{x^2} - 1} \right) = f\left( x \right) \Leftrightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = \frac{1}{{{x^2} - 1}}\) 

\( \Rightarrow \int\limits_0^{\frac{1}{2}} {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = \int\limits_0^{\frac{1}{2}} {\frac{1}{{{x^2} - 1}}} } dx \Leftrightarrow \ln \left| {f\left( x \right)} \right|\left| \begin{array}{l}
^{\frac{1}{2}}\\
_0
\end{array} \right. = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right|\left| \begin{array}{l}
^{\frac{1}{2}}\\
_0
\end{array} \right. \Leftrightarrow \ln \left| {f\left( {\frac{1}{2}} \right)} \right| - \ln \left| e \right| = \frac{1}{2}\left( {\ln \frac{1}{3} - \ln 1} \right)\) 

\( \Leftrightarrow \ln \left| {f\left( {\frac{1}{2}} \right)} \right| - 1 =  - \frac{1}{2}\ln 3 \Leftrightarrow \ln \left| {f\left( {\frac{1}{2}} \right)} \right| = \ln \frac{e}{{\sqrt 3 }} \Leftrightarrow \left| {f\left( {\frac{1}{2}} \right)} \right| = \frac{e}{{\sqrt 3 }} \Rightarrow f\left( {\frac{1}{2}} \right) = \frac{e}{{\sqrt 3 }}\) (do hàm số \(f(x)\) dương)