6.034. $1+\sqrt{1+x \sqrt{x^{2}-24}}=x$.

6.034. $1+\sqrt{1+x \sqrt{x^{2}-24}}=x$.
Solution.

Write the equation as $\sqrt{1+x \sqrt{x^{2}-24}}=x-1$. Squaring both sides of the equation, we get

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 1 + x \sqrt { x ^ { 2 } - 2 4 } = x ^ { 2 } - 2 x + 1 , } \\
{ x - 1 \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \sqrt{x^{2}-24}=x^{2}-2 x, \\
x \geq 1
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
\sqrt{x^{2}-24}=x-2, \\
x \geq 1
\end{array}\right. \\
& \Rightarrow x^{2}-24=x^{2}-4 x+4 \text { or } 4 x=28 ; x=7 .
\end{aligned}
$$

Checking $x=7$ by direct substitution into the original equation, we have:

Answer: $x=7$.