嵌套根式

设该无限嵌套根式的值为 $x>0$。则它满足自相似关系

$$x=\sqrt{2+x}.$$

两边平方得

$$x^2=2+x\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0.$$

取正根，得 $\boxed{x=2}$。

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英文解析

Let the value of this infinitely nested radical be $x>0$. Then it satisfies the self-similar relation

$$x=\sqrt{2+x}.$$

Squaring both sides yields

$$x^2=2+x\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0.$$

Taking the positive root, we obtain $\boxed{x=2}$.