If $f\left(x\right)=\begin{array}{cc}\frac{{72}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}& x\ne 0\end{array}$

$=\begin{array}{cc}k\log 2.\log 3& x=0\end{array}$ is a continuous function then k = 

$\begin{array}{l}f\left(x\right)=\frac{{72}^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}\\ f\left(0\right)=\underset{x\to 0}{Lt}f\left(x\right)\\ f\left(0\right)=\underset{x\to 0}{Lt}\frac{(9^x-1)(8^x-1)}{\sqrt{2}-\sqrt{1+\cos x}}\times \frac{(\sqrt{2}+\sqrt{1+\cos x})}{(\sqrt{2}+\sqrt{1+\cos x})}\end{array}$

$f\left(0\right)=\underset{x\to 0}{Lt}\frac{(9^x-1)(8^x-1)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$

$f\left(0\right)=\underset{x\to 0}{Lt}\frac{\left(\frac{9^x-1}{x}\right)\left(\frac{8^x-1}{x}\right)\text{×}{x}^2}{\frac{2{\sin }^2\frac{x}{2}}{\frac{x^2}{4}}\times \frac{x^2}{4}}$      $\underset{x\to 0}{Lt}(\sqrt{2}+\sqrt{1+\cos x})$

$\begin{array}{l}f\left(0\right)=k\log 2.\log 3=2\log 9.\log 8(\sqrt{2}+\sqrt{2})\\ k.\log 2.\log 3=2.2\sqrt{2}.3\left(\log 2\right).2\left(\log 3\right)\\ k=24\sqrt{2}\end{array}$