Ta có: f(x)+3f(1/x)=x^2
với x=2 ta có:
\(f\left(3\right)+3f\left(\frac{1}{2}\right)=4\)(1)
với x=1/2 ta có:
\(f\left(\frac{1}{2}\right)+3f\left(2\right)=\frac{1}{4}\)(2)
=> \(f\left(\frac{1}{2}\right)=\frac{1}{4}-3f\left(2\right)\)
lấy (1) trừ 2 , ta được:
a\(\left[f\left(2\right)+3f\left(\frac{1}{2}\right)\right]-\left[f\left(\frac{1}{2}\right)+3f\left(2\right)\right]=4-\frac{1}{4}\)
=>\(f\left(2\right)+3f\left(\frac{1}{2}\right)-f\left(\frac{1}{2}\right)-3f\left(2\right)=\frac{15}{4}\)
=>\(2f\left(\frac{1}{2}\right)-2f\left(2\right)=\frac{15}{4}\)
=>\(2\left[f\left(\frac{1}{2}\right)-f\left(2\right)\right]=\frac{15}{4}\)
=> \(f\left(\frac{1}{2}\right)-f\left(2\right)=\frac{15}{8}\)
mà \(f\left(\frac{1}{2}\right)=\frac{1}{4}-3f\left(2\right)\)
=>\(\frac{1}{4}-3f\left(2\right)-f\left(2\right)=\frac{15}{8}\)
=>\(-4f\left(2\right)=-\frac{17}{8}\)
=> \(f\left(2\right)=-\frac{17}{32}\)
vậy...
