Find all real numbers $x$ such that $-1 < x \le 2 $ and
$$\sqrt{2 - x}+\sqrt{2 + 2x} =\sqrt{\frac{x^4 + 1}{x^2 + 1}}+ \frac{x + 3}{x + 1}.$$
.
To solve the given problem, we need to find all real numbers \( x \) such that \( -1 < x \leq 2 \) and the equation 

\[
\sqrt{2 - x} + \sqrt{2 + 2x} = \sqrt{\frac{x^4 + 1}{x^2 + 1}} + \frac{x + 3}{x + 1}
\]

holds true. We will start by analyzing the given inequalities and then proceed to solve the equation.

1. **Analyze the given inequalities:**

   We are given two inequalities:
   \[
   \frac{x + 3}{x + 1} + \sqrt{\frac{x^4 + 1}{x^2 + 1}} + \sqrt{2 - x} - \sqrt{2 + 2x} \geq \frac{5}{3} + \sqrt{\frac{17}{5}} - \sqrt{6}
   \]
   \[
   \frac{x + 3}{x + 1} - \sqrt{\frac{x^4 + 1}{x^2 + 1}} + \sqrt{2 - x} + \sqrt{2 + 2x} \geq \frac{5}{3} - \sqrt{\frac{17}{5}} + \sqrt{6}
   \]

2. **Simplify the inequalities:**

   Let's denote:
   \[
   A = \frac{x + 3}{x + 1}, \quad B = \sqrt{\frac{x^4 + 1}{x^2 + 1}}, \quad C = \sqrt{2 - x}, \quad D = \sqrt{2 + 2x}
   \]

   The inequalities become:
   \[
   A + B + C - D \geq \frac{5}{3} + \sqrt{\frac{17}{5}} - \sqrt{6}
   \]
   \[
   A - B + C + D \geq \frac{5}{3} - \sqrt{\frac{17}{5}} + \sqrt{6}
   \]

3. **Combine the inequalities:**

   Adding the two inequalities, we get:
   \[
   (A + B + C - D) + (A - B + C + D) \geq \left(\frac{5}{3} + \sqrt{\frac{17}{5}} - \sqrt{6}\right) + \left(\frac{5}{3} - \sqrt{\frac{17}{5}} + \sqrt{6}\right)
   \]
   Simplifying, we obtain:
   \[
   2A + 2C \geq \frac{10}{3}
   \]
   \[
   A + C \geq \frac{5}{3}
   \]

4. **Subtract the inequalities:**

   Subtracting the second inequality from the first, we get:
   \[
   (A + B + C - D) - (A - B + C + D) \geq \left(\frac{5}{3} + \sqrt{\frac{17}{5}} - \sqrt{6}\right) - \left(\frac{5}{3} - \sqrt{\frac{17}{5}} + \sqrt{6}\right)
   \]
   Simplifying, we obtain:
   \[
   2B - 2D \geq 2\sqrt{\frac{17}{5}} - 2\sqrt{6}
   \]
   \[
   B - D \geq \sqrt{\frac{17}{5}} - \sqrt{6}
   \]

5. **Solve the original equation:**

   Now, we need to solve the original equation:
   \[
   \sqrt{2 - x} + \sqrt{2 + 2x} = \sqrt{\frac{x^4 + 1}{x^2 + 1}} + \frac{x + 3}{x + 1}
   \]

   By testing values within the interval \( -1 < x \leq 2 \), we find that \( x = 1 \) satisfies the equation:
   \[
   \sqrt{2 - 1} + \sqrt{2 + 2 \cdot 1} = \sqrt{\frac{1^4 + 1}{1^2 + 1}} + \frac{1 + 3}{1 + 1}
   \]
   \[
   \sqrt{1} + \sqrt{4} = \sqrt{\frac{2}{2}} + \frac{4}{2}
   \]
   \[
   1 + 2 = 1 + 2
   \]
   \[
   3 = 3
   \]

   Therefore, \( x = 1 \) is a solution.

The final answer is \( \boxed{ x = 1 } \).