Ta có : f(x) + 2f(\(\dfrac{1}{x}\))=x2 =>f(x)=x2 - 2f(\(\dfrac{1}{x}\))
=> f(2)=22- 2f(\(\dfrac{1}{2}\))=4- 2f(\(\dfrac{1}{2}\)) (1)
Ta lại có: f(\(\dfrac{1}{2}\))= (\(\dfrac{1}{2}\))2-2f(\(\dfrac{1}{\dfrac{1}{2}}\))=\(\dfrac{1}{4}\)-2f(2)
Thay f(\(\dfrac{1}{2}\))=\(\dfrac{1}{4}\)-2f(2) vào (1) ta có:
f(2)= 4- 2.[\(\dfrac{1}{4}\)-2f(2)] => f(2)=4-\(\dfrac{1}{2}\)+4f(2)
=> -3.f(2)=\(\dfrac{7}{2}\)=> f(2)=-\(\dfrac{7}{6}\)
bạn nhớ chọn mk nhé