We are given:

$$
\(y=\frac{p}{q} \quad \text { and } \quad x=\frac{p}{p+q}\)
$$


We need to determine which of the following options are true:
- (A) $\(x=\frac{y}{y+1}\)$
- (B) $\(x=\frac{y}{y-1}\)$
- (C) $\(y=\frac{x}{x-1}\)$
- (D) $\(y=\frac{x}{1-x}\)$

Step 1: Express $p$ in Terms of $y$ and $q$
From $\(y=\frac{p}{q}\)$, we can express $p$ as:

$$
\(p=y q\)
$$


Step 2: Substitute $p$ into the Expression for $x$
Given $\(x=\frac{p}{p+q}\)$, substitute $\(p=y q\)$ :

$$
\(x=\frac{y q}{y q+q}=\frac{y q}{q(y+1)}=\frac{y}{y+1}\)
$$


This matches Option (A).

Step 3: Solve for $y$ in Terms of $x$
Starting from $\(x=\frac{y}{y+1}\)$ :
1. Multiply both sides by $\(y+1\)$ :

$$
\(x(y+1)=y\)
$$

2. Expand and rearrange:

$$
\(\begin{gathered}
x y+x=y \\
x=y-x y \\
x=y(1-x)
\end{gathered}\)
$$

3. Solve for $y$ :

$$
\(y=\frac{x}{1-x}\)
$$


This matches Option (D).
Step 4: Verify Options (B) and (C)
- Option (B): $\(x=\frac{y}{y-1}\)$
- From Step 2, we have $\(x=\frac{y}{y+1}\)$, which contradicts $\(\frac{y}{y-1}\)$. Hence, false.
- Option (C): $\(y=\frac{x}{x-1}\)$
- From Step 3, we derived $\(y=\frac{x}{1-x}\)$, which is not equal to $\(\frac{x}{x-1}\)$. Hence, false.

Conclusion
Only Options (A) and (D) are true.
Final Answer
\(\boxed\{A, D\}\)