Because \(x,y\in\left\{1;2;...;40\right\}\) and we must find \(Max\dfrac{x+y}{x-y}\) 

=> x - y = 1 and \(max\left(x+y\right)\)

=> We have: \(\dfrac{40+39}{40-39}=\dfrac{79}{1}=79\) satisfy the operation.

=> 

Let \(\dfrac{X+Y}{X-Y}_{MAX}\) so X-Y must in range 1 - the minimum range (X-Y mustn't in range 0 because it is denominator).

So from 1 to 40 we can choose \(\left\{{}\begin{matrix}X=21\\Y=20\end{matrix}\right.\Leftrightarrow\dfrac{X+Y}{X-Y}=\dfrac{21+20}{21-20}=\dfrac{41}{1}=41\)

So the largest value that \(\dfrac{X+Y}{X-Y}\) can have is 41.