The number of real solutions of the equation  $x^9-\frac{288}{5}{x}^5+2304x+2=0$  is equal to 

 $$\begin{gathered} f\left(x\right)=x^9-\frac{288}{5}{x}^5+2304x+2=0 \\ \begin{array}{l}{f}^1\left(x\right)=0\Rightarrow 9x^8-288x^4+2304=0\\ \Rightarrow x^8-32x^4+256=0\\ \Rightarrow {\left(x^4-16\right)}^2=0\Rightarrow x=\pm 2\end{array} \end{gathered}$$
 Observe that  $f\left(2\right)>0 \therefore f\left(x\right)=0$  has only one real root