The exercise is asking for the negation of the following propositions:

1. Proposition r: $\forall x \in \mathbb{R}, -1 \leq \frac{2x}{x^2+1} \leq 1.$

2. Proposition p: $\exists (x, y, z) \in \mathbb{R}^3 : \frac{x+1}{7} = \frac{y-5}{2} = \frac{z}{13}.$

3. Proposition q: $\forall x \in \mathbb{R}, x^2 \in \mathbb{Q} \Rightarrow x \in \mathbb{Q}.$

Now, let's provide the negations for each:

Negation of r:

The original statement is: $\forall x \in \mathbb{R}, -1 \leq \frac{2x}{x^2 + 1} \leq 1.$ The negation would be: $\exists x \in \mathbb{R} \text{ such that } \left( \frac{2x}{x^2 + 1} < -1 \right) \text{ or } \left( \frac{2x}{x^2 + 1} > 1 \right).$ This means there exists at least one $x$ where the inequality is not satisfied.

Negation of p:

The original statement is: $\exists (x, y, z) \in \mathbb{R}^3 : \frac{x+1}{7} = \frac{y-5}{2} = \frac{z}{13}.$ The negation would be: $\forall (x, y, z) \in \mathbb{R}^3, \frac{x+1}{7} \neq \frac{y-5}{2} \text{ or } \frac{y-5}{2} \neq \frac{z}{13}.$ This means for all possible triples $(x, y, z)$, the equality does not hold.

Negation of q:

The original statement is: $\forall x \in \mathbb{R}, x^2 \in \mathbb{Q} \Rightarrow x \in \mathbb{Q}.$ The negation would be: $\exists x \in \mathbb{R} \text{ such that } x^2 \in \mathbb{Q} \text{ and } x \notin \mathbb{Q}.$ This means there exists some real number $x$ such that $x^2$ is rational, but $x$ itself is not rational.

Do you want further details or have any questions?

Here are 5 related questions you might explore:

1. How do logical quantifiers affect negation?
2. Can you find an example where $\frac{2x}{x^2 + 1}$ exceeds the given bounds?
3. What is the geometric meaning of the system of equations in proposition $p$?
4. Can you think of a real number whose square is rational but the number itself is not?
5. How does negating a mathematical implication $P \Rightarrow Q$ differ from negating conjunctions or disjunctions?

Tip: To negate a statement involving "for all" ($\forall$) or "there exists" ($\exists$), always remember that these switch roles during negation: $\forall$ becomes $\exists$ and $\exists$ becomes $\forall$.