<s>[INST] Two Problem: find $\max, \min$; number theory: find $x, y$ 
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*Find $x, y \in \mathbb{N}$ such that $$\left.\frac{x^2+y^2}{x-y}~\right|~ 2010$$

*Find max and min of $\sqrt{x+1}+\sqrt{5-4x}$ (I know $\max = \frac{3\sqrt{5}}2,\, \min = \frac 3 2$)

 [/INST] The second problem can be solved using routine calculus. There are also various non-calculus solutions. These in principle involve only basic machinery from algebra or trigonometry, but are in practice more complicated than the calculus solution.
The derivative of our function is 
$$\frac{1}{2}\left(\frac{1}{\sqrt{x+1}}-\frac{4}{\sqrt{5-4x}}\right).$$
When you set this equal to $0$, and manipulate a little, you will get a linear equation for $x$. Our candidates are the solution of this equation, and the natural endpoints $x=-1$ and $x=\frac{5}{4}$. So we have $3$ candidates.  Substitution will determine the winner and the loser. 
For the first problem, it is not clear whether you want one solution, or a characterization of all solutions. We will find a couple of solutions, and you can do the rest.
Let's see if we can make the divisor $\frac{x^2+y^2}{x-y}$ of $2010$ equal to $2010$.  We get the equation $\frac{x^2+y^2}{x-y}=2010$. Rewrite this equation as $x^2+y^2=2010(x-y)$, and then, completing the square, as
$$(x-1005)^2+(y+1005)^2=2(1005)^2.$$
So we want to express $2(1005)^2$ as a sum of two squares. We also want to make sure our solution produces positive $x$ and $y$.
Note that $2(1005)^2=50(201)^2= (201)^2+ 7^2(201)^2$. 
Set $x-1005=201$ and $y+1005=7(201)$, and solve.
To find other solutions, repeat the calculation with divisors $d$ of $2010$ different from $2010$. By if necessary interchanging the roles of $x$ and $y$, we can confine attention to positive divisors. 
Note that $2010$ has few divisors. The work is also made easier by the fact that if $q$ is a prime of the form $4k+3$, and the prime factorization of $m$ contains $q$ to an odd power, then the equation $s^2+t^2=m$ has no solutions. And if the prime factorization of $m$ contains $q$ to an even power, and $s^2+t^2=m$, then $q$ divides $s$ and $t$. Since $2010=(2)93)(5)(67)$, two of the prime factors of $m$ are of the form $4k+3$. This greatly simplifies the search for solutions. 
The easiest thing to deal with is the divisor $10$. We end up looking at the equation $(x-5)^2+(y+5)^2= 50$, which gives $x=6$, $y=2$.  
Slightly harder is the odd divisor $5$. When we complete the square, we get
$(x-5/2)^2+(y+5/2)^2=25/2$. Multiply through by $4$.    
There are not many other cases to consider. Your turn!
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