6. (9th "Hope Cup" Invitational Competition Question) Let $\alpha, \beta$ be the roots of the equations $\log _{2} x+x+2=0$ and $2^{x}+x+2=0$ respectively, then the value of $\alpha+\beta$ is equal to

保留源文本的换行和格式，翻译结果如下：

6. (9th "Hope Cup" Invitational Competition Question) Let $\alpha, \beta$ be the roots of the equations $\log _{2} x+x+2=0$ and $2^{x}+x+2=$ 0 respectively, then the value of $\alpha+\beta$ is equal to
6. Fill in 2. Reason: It is easy to know that the function $f(x)=\log _{2} x+x+2$ is monotonically increasing in the interval $(0,+\infty)$. When $\log _{2} x+x+2=0$, we get $x=2^{-x-2}$. By substituting $-x-2$ for $x$, we get $-x+2=2^{x}$, which is equivalent to $x=\log _{2}(-x-2)$, thus $\log _{2}(-x-2)+$ $2=0$.

Let $\alpha$ be the root of the equation $x=2^{-x-2}$, i.e., $f(\alpha)=0$, and let $\beta$ be the root of $-x-2=2^{x}$, i.e., the root of $2^{x}+x+2=0$. From $*$, we know that $f(-\beta-2)=0$, so $f(\alpha)=f(-\beta-2)$, which means $\alpha+\beta=-2$.