4. 42 Find the positive root of the equation $x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}$, and prove that there is only one positive root.
[Solution] (1) If $x=2$, then
$$
\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2}}}}=2 \text{.}
$$
$\therefore 2$ is the positive root of the equation.
(2) If $x>2$, let $x=2+\alpha(\alpha>0)$,

then $\because \quad(2+\alpha)^{2}=4+4 \alpha+\alpha^{2}>4+\alpha>4$,
$$
\therefore \quad x^{2}>2+x>4, x>\sqrt{2+x}>2 \text{.}
$$

Similarly, we can get:
$$
\begin{aligned}
& \sqrt{2+x}>\sqrt{2+\sqrt{2+x}}>2, \\
& \sqrt{2+\sqrt{2+x}}>\sqrt{2+\sqrt{2+\sqrt{2+x}}}>2, \\
& \cdots \cdots \cdots \cdots \\
\therefore \quad & x>\sqrt{2+x}>\sqrt{2+\sqrt{2+x}}>\sqrt{2+\sqrt{2+\sqrt{2+x}}}
\end{aligned}
$$

Thus, when $x>2$, it cannot be the root of the equation.
(3) If $0<x<2$, let $x=2-\alpha(\alpha>0)$.
then $\because \quad(2-\alpha)^{2}=4-4 \alpha+\alpha^{2}=4-\alpha(4-\alpha)$
and $\quad 4-\alpha>1$,
$$
\begin{array}{ll}
\therefore \quad & 4>4-\alpha>4-\alpha(4-\alpha), \\
& 4>2+x>x^{2}, \\
& 2>\sqrt{2+x}>x .
\end{array}
$$

Similarly, we can get:
$$
\begin{array}{l}
2>\sqrt{2+\sqrt{2+x}}>\sqrt{2+x} \text{,} \\
2>\sqrt{2+\sqrt{2+\sqrt{2+x}}}>\sqrt{2+\sqrt{2+x}} \text{,} \\
\text{...........} \\
\therefore \quad \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}>\sqrt{2+\sqrt{2+\sqrt{2+x}}} \\
>\sqrt{2+\sqrt{2+x}}>\sqrt{2+x}>x \text{.} \\
\end{array}
$$

Thus, when $0<x<2$, it also cannot be the root of the equation. Therefore, this equation has only one positive root, which is 2.