Tham khảo:

refer

refer

hàm số f(x) xác định với mọi x thỏa mãn \(f\left(x\right)+2f\left(\frac{1}{x}\right)=x^2\)nên:

+) x = 3 thì \(f\left(3\right)+2f\left(\frac{1}{3}\right)=\frac{1}{9}\Rightarrow2f\left(3\right)+4f\left(\frac{1}{3}\right)=\frac{2}{9}\)(1)

+) x = \(\frac{1}{3}\)thì \(f\left(\frac{1}{3}\right)+2f\left(3\right)=9\)(2)

Lấy (1) - (2) ta được: \(3f\left(\frac{1}{3}\right)=\frac{-79}{9}\)

\(\Rightarrow f\left(\frac{1}{3}\right)=\frac{-79}{27}\)

Làm ngược, sửa:))

+) Nếu x = 3 thì \(f\left(3\right)+2f\left(\frac{1}{3}\right)=9\Rightarrow2f\left(3\right)+4f\left(\frac{1}{3}\right)=18\)(1)

+) Nếu x = \(\frac{1}{3}\) thì \(f\left(\frac{1}{3}\right)+2f\left(3\right)=\frac{1}{9}\)(2)

Lấy (1) - (2) ta được: \(3f\left(\frac{1}{3}\right)=\frac{161}{9}\)

\(\Rightarrow f\left(\frac{1}{3}\right)=\frac{161}{7}\)

\(f\left(x\right)+2f\left(\frac{1}{x}\right)=x^2\) (1)

\(\Rightarrow f\left(\frac{1}{x}\right)+2f\left(x\right)=\frac{1}{x^2}\Rightarrow2f\left(\frac{1}{x}\right)+4f\left(x\right)=\frac{2}{x^2}\) (2)

Trừ (2) cho (1): \(3f\left(x\right)=\frac{2}{x^2}-x^2\Rightarrow f\left(x\right)=\frac{2}{3x^2}-\frac{1}{3}x^2\)

\(\Rightarrow f\left(2019\right)=\frac{2}{3.2019^2}-\frac{1}{3}.2019^2\)

Chọn đáp án C.

Chọn D.

Đặt  u   =   x + 1 d v   =   f ' ( x )   d x ⇒ d u   =   d x v   =   ∫ f ' ( x ) d x

⇔ 10 = 2f(1) – f(0) – I ⇔ 10 = 2 – I ⇔ I = -8.