## Task 1 - 311241

Let

$$
x=e^{0.000009}-e^{0.000007}+e^{0.000002}-e^{0.000001} ; \quad y=e^{0.000008}-e^{0.000005}
$$

Determine whether $x=y$ or $x>y$ or $x<y$.
We set

$$
z=e^{0.000001} \approx 1.000001>1
$$

Then we have

$$
\begin{gathered}
x-y=z^{9}-z^{8}-z^{7}+z^{5}+z^{2}-z \\
x-y=z\left(z^{8}-z^{7}-z^{6}+z^{4}+z-1\right) \\
x-y=z(z-1)\left(z^{7}-z^{4}(z+1)+1\right) \\
x-y=z(z-1)\left(z^{7}-z^{5}-z^{4}+1\right) \\
x-y=z(z-1)\left(z^{2}-1\right)\left(z^{5}-z^{2}-1\right)
\end{gathered}
$$

Since $z>1$, the factors $z, (z-1)$, and $\left(z^{2}-1\right)$ are all greater than zero. Since $\left(z^{5}-z^{2}-1\right) \approx-1<0$, it follows that $x-y<0$ or $x<y$.