simplify (1)

Let $\frac xy=a,\dfrac{\frac{a+2}{a+1}+a}{a+2-\frac{a}{a+1}}=\dfrac{\frac{a+2}{a+1}+a}{a+\frac{a+2}{a+1}}=1$

Assume $x = 2017 = a$ and $y = \dfrac 1{2017} = \dfrac 1a$ . So $\dfrac xy = a^2$ .

$\begin{aligned} \left(\frac{\frac xy+2}{\frac xy+1}+\frac xy \right)\div \left(\frac xy+2-\frac{\frac xy}{\frac xy+1}\right) & = \left(\dfrac{a^2 + 2}{a^2 +1} + a^2 \right) \div \left(a^2 + 2 - \dfrac{a^2}{a^2 + 1} \right) \\ & = \dfrac{a^2+2 + a^2(a^2 +1)}{a^2 + 1} \times \dfrac{a^2 +1}{(a^2 + 1)(a^2 +2) - a^2} \\ & = \dfrac{a^2 + 2 + a^4 + a^2}{a^4 + 2a^2 + a^2 + 2 - a^2} \\ & = \dfrac{a^4 + 2a^2 + 2}{a^4 + 2a^2 + 2} \\ & = \boxed 1 \\ \end{aligned}$

$\begin{aligned} Q & = \left(\frac{\frac{x}{y}+2}{\frac{x}{y}+1}+\frac{x}{y}\right)\div \left(\frac{x}{y}+2-\frac{\frac{x}{y}}{\frac{x}{y}+1}\right) & \small \color{#3D99F6} \text{Let }v = \frac xy \\ & = \frac {\frac {v+2}{v+1}+v}{v+2-\frac v{v+1}} & \small \color{#3D99F6} \text{Multiply up and down by }(v+1) \\ & = \frac {v+2+v(v+1)}{(v+2)(v+1)-v} \\ & = \frac {v^2+2v+2}{v^2+2v+2} \\ & = \boxed{1} \end{aligned}$