Tiếp bài 5 giúp
perfectstrong.
$f({x^3} - {y^3}) = {x^2}f(x) - {y^2}f(y)$ (1)
Trong (1) cho $y = 0 \Rightarrow f\left( {{x^3}} \right) = {x^2}f\left( x \right)\,\,\,\forall x \in R$. Mặt khác ta có:
$f\left( {{x^3} - {y^3}} \right) = f\left( {{x^3}} \right) - f\left( {{y^3}} \right),\,\,\forall x,y \in R \Rightarrow f\left( {x - y} \right) = f\left( x \right) - f\left( y \right)\,\,\,\,\,(2)$
Trong (2) cho $x = y \Rightarrow f\left( 0 \right) = 0;\,\,x = 0 \Rightarrow f\left( { - y} \right) = - f\left( y \right),\,\,\forall y \in R$.
Khi đó (2) viết thành: $f\left( {x + y} \right) = f\left( x \right) + f\left( y \right),\,\,\,\forall x,y \in R\,\,\,\,(3)$
Hàm cộng tính nên $f\left( {kx} \right) = kf\left( x \right),\,\,\forall x \in R,\forall k \in Q$. Từ đó ta được:
$\left\{ \begin{array}{l}f\left( {{x^3} + {y^3}} \right) = {x^2}f\left( x \right) + {y^2}f\left( y \right),\,\,\forall x,y \in R\\f\left( {{x^3}} \right) = {x^2}f\left( x \right),\,\,\forall x \in R\\f\left( {x + y} \right) = f\left( x \right) + f\left( y \right),\,\,\,\forall x,y \in R\\f\left( {kx} \right) = kf\left( x \right),\,\,\forall x \in R,\forall k \in Q\end{array} \right.$
Khi đó: $f\left( {{{\left( {x - 1} \right)}^3} + {{\left( {x + 1} \right)}^3}} \right) = f\left( {2{x^3} + 6x} \right) = f\left( {2{x^3}} \right) + f\left( {6x} \right) = 2{x^2}f\left( x \right) + 6f\left( x \right),\forall x \in R$
Do đó: $2{x^2}f\left( x \right) + 6f\left( x \right) = \left( {2{x^2} + 2} \right)f\left( x \right) + 4xf\left( 1 \right),\forall x \in R \Leftrightarrow f\left( x \right) = f\left( 1 \right)x,\forall x \in R$.