1. Given quadratic and its roots

The equation is $x^2 + 5\sqrt{2}\,x + 10 = 0.$ Let its roots be $\alpha, \beta$ with $\alpha > \beta$.

By Vieta's formulas, $\alpha + \beta = -5\sqrt{2}, \qquad \alpha\beta = 10.$

Also, $P_n = \alpha^n - \beta^n.$

2. Simplify the required expression

We need to find $E=\frac{P_{17}P_{20}+5\sqrt{2}\,P_{17}P_{19}}{P_{18}P_{19}+5\sqrt{2}\,P_{18}^2}.$

Factor common terms from numerator and denominator: $E = \frac{P_{17}(P_{20}+5\sqrt{2}P_{19})}{P_{18}(P_{19}+5\sqrt{2}P_{18})}.$

So it remains to understand terms of the form $P_{n+1}+5\sqrt{2}P_n.$

3. Derive a recurrence for $P_n$

Since $\alpha, \beta$ are roots of $x^2+5\sqrt{2}x+10=0,$ each root satisfies $r^2+5\sqrt{2}r+10=0 \quad \Rightarrow \quad r^{n+2}+5\sqrt{2}r^{n+1}+10r^n=0.$

Applying this for $r=\alpha$ and $r=\beta$, and subtracting, we get $P_{n+2}+5\sqrt{2}P_{n+1}+10P_n=0.$

Hence, $P_{n+2}+5\sqrt{2}P_{n+1}=-10P_n.$

Replacing $n$ by $n-1$, $P_{n+1}+5\sqrt{2}P_n=-10P_{n-1}.$

4. Use this identity in the expression

For the numerator part: $P_{20}+5\sqrt{2}P_{19}=-10P_{18}.$

For the denominator part: $P_{19}+5\sqrt{2}P_{18}=-10P_{17}.$

Substitute into $E$: $E = \frac{P_{17}(-10P_{18})}{P_{18}(-10P_{17})}.$

Therefore, $E=1.$

5. Final answer

$\boxed{1}$

The derived answer matches the stored correct answer.