For each real value of $p$, find all real solutions of the equation $\sqrt{x^2 - p}+2\sqrt{x^2-1} = x$.
1. **Initial Conditions and Substitution:**
   Given the equation:
   \[
   \sqrt{x^2 - p} + 2\sqrt{x^2 - 1} = x
   \]
   Note that for the square roots to be defined, we must have \(x^2 \geq p\) and \(x^2 \geq 1\). Since \(x^2 \geq 1\), it follows that \(x \geq 1\).

   Let \(t = \sqrt{x^2 - 1}\). Then \(t \geq 0\) and \(x^2 = t^2 + 1\). Substituting \(t\) into the original equation, we get:
   \[
   \sqrt{t^2 + 1 - p} + 2t = \sqrt{t^2 + 1}
   \]

2. **Isolate the Square Root:**
   Isolate \(\sqrt{t^2 + 1 - p}\):
   \[
   \sqrt{t^2 + 1 - p} = \sqrt{t^2 + 1} - 2t
   \]

3. **Square Both Sides:**
   Square both sides to eliminate the square root:
   \[
   t^2 + 1 - p = (\sqrt{t^2 + 1} - 2t)^2
   \]
   Expanding the right-hand side:
   \[
   t^2 + 1 - p = t^2 + 1 - 4t\sqrt{t^2 + 1} + 4t^2
   \]
   Simplifying, we get:
   \[
   -p = 4t^2 - 4t\sqrt{t^2 + 1}
   \]
   \[
   p = 4t\sqrt{t^2 + 1} - 4t^2
   \]

4. **Isolate the Square Root Again:**
   Isolate \(4t\sqrt{t^2 + 1}\):
   \[
   4t\sqrt{t^2 + 1} = 4t^2 + p
   \]

5. **Square Both Sides Again:**
   Square both sides again:
   \[
   (4t\sqrt{t^2 + 1})^2 = (4t^2 + p)^2
   \]
   \[
   16t^2(t^2 + 1) = 16t^4 + 8pt^2 + p^2
   \]
   Simplifying, we get:
   \[
   16t^4 + 16t^2 = 16t^4 + 8pt^2 + p^2
   \]
   \[
   16t^2 = 8pt^2 + p^2
   \]
   \[
   16t^2 - 8pt^2 = p^2
   \]
   \[
   t^2(16 - 8p) = p^2
   \]
   \[
   t^2 = \frac{p^2}{16 - 8p}
   \]

6. **Find \(x\):**
   Recall that \(x^2 = t^2 + 1\):
   \[
   x^2 = 1 + \frac{p^2}{16 - 8p}
   \]
   \[
   x = \sqrt{1 + \frac{p^2}{16 - 8p}}
   \]

7. **Simplify \(x\):**
   Simplify the expression for \(x\):
   \[
   x = \frac{4 - p}{\sqrt{16 - 8p}}
   \]

8. **Check Conditions:**
   We need to check the conditions for \(t\) and \(p\):
   - For \(t \geq 0\), we have \(\sqrt{t^2 + 1} \geq 2t\):
     \[
     t^2 + 1 \geq 4t^2
     \]
     \[
     1 \geq 3t^2
     \]
     \[
     t^2 \leq \frac{1}{3}
     \]
     \[
     \frac{p^2}{16 - 8p} \leq \frac{1}{3}
     \]
     \[
     3p^2 \leq 16 - 8p
     \]
     \[
     3p^2 + 8p - 16 \leq 0
     \]
     Solving the quadratic inequality:
     \[
     (3p - 4)(p + 4) \leq 0
     \]
     \[
     -4 \leq p \leq \frac{4}{3}
     \]

   - For \(4t^2 + p \geq 0\):
     \[
     4 \cdot \frac{p^2}{16 - 8p} + p \geq 0
     \]
     Simplifying, we find that \(0 \leq p \leq 4\).

   Combining these conditions, we get:
   \[
   0 \leq p \leq \frac{4}{3}
   \]

The final answer is \( \boxed{ x = \frac{4 - p}{\sqrt{16 - 8p}} } \) for \(0 \leq p \leq \frac{4}{3}\).