7. Calculate with an accuracy of one-tenth the value of the expression $\sqrt{86+41 \sqrt{86+41 \sqrt{86+\ldots}}}$
Solution. Consider the strictly increasing sequence of values:

$$
\sqrt{86}, \sqrt{86+41 \sqrt{86}}, \sqrt{86+41 \sqrt{86+41 \sqrt{86}}}, \ldots
$$

If this sequence is bounded above, then the value of $F$ is the least upper bound, and thus $F$ is a real number. Therefore, it is sufficient to prove the boundedness of the given sequence. We will prove that this sequence is bounded above by the number 43. Indeed,

$$
\begin{aligned}
\sqrt{86}<43, \Rightarrow & 41 \sqrt{86}<41 \cdot 43 \Rightarrow 86+41 \sqrt{86}<86+41 \cdot 43=43^{2} \Rightarrow \Rightarrow \sqrt{86+41 \sqrt{86}} \\
& <43 \text { and so on. }
\end{aligned}
$$

It is clear from the condition that $F$ is the positive root of the equation $F^{2}=86+41 F$. From this, we find $F=43$.

Answer: 43.