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Let $\mathrm{a}>0$ be a root of the equation $2 x^2+x-2=0$. If $\lim _{x \rightarrow \frac{1}{\mathrm{a}}} \frac{16\left(1-\cos \left(2+x-2 x^2\right)\right)}{(1-\mathrm{a} x)^2}=\alpha+\beta \sqrt{17}$, where $\alpha, \beta \in Z$, then $\alpha+\beta$ is equal to_______
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170
$\begin{aligned} & \lim _{x \rightarrow \frac{1}{a}} 16 \cdot \frac{\left(1-\cos 2\left(x-\frac{1}{a}\right)\left(x-\frac{1}{b}\right)\right)}{4\left(x-\frac{1}{b}\right)^2} \times \frac{4\left(x-\frac{1}{b}\right)^2}{a^2\left(x-\frac{1}{a}\right)^2} \\ & =16 \times \frac{2}{a^2}\left(\frac{1}{a}-\frac{1}{b}\right)^2 \\ & =\frac{32}{a^2}\left(\frac{17}{4}\right)=\frac{17.8}{a^2}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^2} \\ & =\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}} \\ & =\frac{136}{256}(18+2 \sqrt{7}) \cdot 16 \\ & =153+17 \sqrt{17}=\alpha+\beta \sqrt{17} \\ & \alpha+\beta=153+17=170\end{aligned}$
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