Find all real $x$ that satisfy the equation $$\frac{1}{x+1}+\frac{1}{x+2}=\frac{1}{x}$$

[i]2015 CCA Math Bonanza Lightning Round #2.2[/i]
To solve the equation 
\[
\frac{1}{x+1} + \frac{1}{x+2} = \frac{1}{x},
\]
we will first find a common denominator for the left-hand side and then simplify the equation.

1. **Combine the fractions on the left-hand side:**
   \[
   \frac{1}{x+1} + \frac{1}{x+2} = \frac{(x+2) + (x+1)}{(x+1)(x+2)} = \frac{2x+3}{(x+1)(x+2)}.
   \]

2. **Set the combined fraction equal to the right-hand side:**
   \[
   \frac{2x+3}{(x+1)(x+2)} = \frac{1}{x}.
   \]

3. **Cross-multiply to eliminate the fractions:**
   \[
   x(2x+3) = (x+1)(x+2).
   \]

4. **Expand both sides:**
   \[
   2x^2 + 3x = x^2 + 3x + 2.
   \]

5. **Simplify the equation by subtracting \(x^2 + 3x + 2\) from both sides:**
   \[
   2x^2 + 3x - (x^2 + 3x + 2) = 0,
   \]
   \[
   2x^2 + 3x - x^2 - 3x - 2 = 0,
   \]
   \[
   x^2 - 2 = 0.
   \]

6. **Solve for \(x\):**
   \[
   x^2 = 2,
   \]
   \[
   x = \pm \sqrt{2}.
   \]

7. **Check for extraneous solutions:**
   We need to ensure that \(x = \sqrt{2}\) and \(x = -\sqrt{2}\) do not make any denominators zero in the original equation.
   - For \(x = \sqrt{2}\):
     \[
     x+1 = \sqrt{2} + 1 \neq 0, \quad x+2 = \sqrt{2} + 2 \neq 0, \quad x = \sqrt{2} \neq 0.
     \]
   - For \(x = -\sqrt{2}\):
     \[
     x+1 = -\sqrt{2} + 1 \neq 0, \quad x+2 = -\sqrt{2} + 2 \neq 0, \quad x = -\sqrt{2} \neq 0.
     \]

Since neither solution makes any denominator zero, both solutions are valid.

The final answer is \(\boxed{\pm \sqrt{2}}\).