Ta có: \(x\cdot f\left(x+2\right)=\left(x^2-9\right)\cdot f\left(x\right)\left(1\right)\)
Thay x=3 vào (1), ta được:
\(3\cdot f\left(3+2\right)=\left(3^2-9\right)\cdot f\left(3\right)\)
=>\(3\cdot f\left(5\right)=0\)
=>f(5)=0
Thay x=7 vào (1), ta được:
\(7\cdot f\left(7+2\right)=\left(7^2-9\right)\cdot f\left(7\right)\)
=>\(7\cdot f\left(9\right)=40\cdot f\left(7\right)\)
=>\(f\left(9\right)=\frac{40}{7}\cdot f\left(7\right)\)
f(5)+f(9)
\(=0+\frac{40}{7}\cdot f\left(7\right)=\frac{40}{7}f\left(7\right)\)