If $a+\alpha=1, b+\beta=2$ and $a f(x)+\alpha f\left(\frac{1}{x}\right)=b x+\frac{\beta}{x}, x \neq 0$, then the value of expression $\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}$ is $\_\_\_\_$ .
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