If $x$  is real, then the maximum value of $\frac{3x^2+9x+17}{3x^2+9x+7}$  is            

Let $y=\frac{3x^2+9x+17}{3x^2+9x+7}$ 

$\Rightarrow y\left(3x^2+9x+7\right)=3x^2+9x+17$

$\Rightarrow 3x^{2}y+9xy+7y-3x^2-9x-17$

$\Rightarrow 3\left(y-1\right)x^2+9\left(y-1\right)x+\left(7y-17\right)=0$

Here $a=3(y-1),b=9(y-1),c=7y-17$

$\Delta \ge 0\Rightarrow 81{\left(y-1\right)}^2-12\left(y-1\right)\left(7y-17\right)\ge 0$      $\left(\because b^2-4ac\ge 0;x\text{ is real}\right)$

           $\Rightarrow -y^2+42y-41\ge 0$

           $\Rightarrow y^2-42y+41\le 0$

           $\Rightarrow \left(y-1\right)\left(y-41\right)\le 0$

           $\Rightarrow y\in \left[1,41\right]$

$\therefore$ The maximum value of given equation $y_{\max }=41$