\section*{Exercise 4 - 131044}

Determine whether the number

\[
x=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}
\]

is positive, negative, or equal to zero!
}

It is

\[
\begin{aligned}
x+\sqrt{2}=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}} & =\frac{(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}) \cdot(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}})}{\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}}= \\
& =\frac{4+\sqrt{7}-(4-\sqrt{7})}{\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}}
\end{aligned}
\]

Then it is

\[
\begin{gathered}
x=0 \Leftrightarrow x+\sqrt{2}=\sqrt{2} \Leftrightarrow \frac{2 \sqrt{7}}{\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}}=\sqrt{2} \\
\Leftrightarrow 2 \sqrt{7}=\sqrt{2} \cdot(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}) \Leftrightarrow \sqrt{2} \cdot \sqrt{7}=(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}) \\
\Leftrightarrow 14=(4+\sqrt{7})+(4-\sqrt{7})+2 \sqrt{4+\sqrt{7}} \cdot \sqrt{4-\sqrt{7}} \\
\Leftrightarrow 6=2 \sqrt{(4+\sqrt{7}) \cdot(4-\sqrt{7})} \Leftrightarrow 3=\sqrt{16-7}=\sqrt{9}=3
\end{gathered}
\]

which is a true statement. Therefore, \(x=0\).