If $x \in R-\{-7\}$, determine the smallest value of the expression
$$\frac{2x^2 + 98}{(x + 7)^2}$$
To determine the smallest value of the expression 
\[ \frac{2x^2 + 98}{(x + 7)^2}, \]
we will analyze the given expression and use calculus to find its minimum value.

1. **Rewrite the expression:**
   \[ f(x) = \frac{2x^2 + 98}{(x + 7)^2}. \]

2. **Simplify the expression:**
   \[ f(x) = \frac{2x^2 + 98}{x^2 + 14x + 49}. \]

3. **Find the critical points by taking the derivative:**
   Let \( f(x) = \frac{2x^2 + 98}{x^2 + 14x + 49} \).
   We use the quotient rule for differentiation:
   \[ f'(x) = \frac{(2x^2 + 98)'(x^2 + 14x + 49) - (2x^2 + 98)(x^2 + 14x + 49)'}{(x^2 + 14x + 49)^2}. \]

4. **Compute the derivatives:**
   \[ (2x^2 + 98)' = 4x, \]
   \[ (x^2 + 14x + 49)' = 2x + 14. \]

5. **Substitute the derivatives into the quotient rule:**
   \[ f'(x) = \frac{4x(x^2 + 14x + 49) - (2x^2 + 98)(2x + 14)}{(x^2 + 14x + 49)^2}. \]

6. **Simplify the numerator:**
   \[ 4x(x^2 + 14x + 49) = 4x^3 + 56x^2 + 196x, \]
   \[ (2x^2 + 98)(2x + 14) = 4x^3 + 28x^2 + 196x + 1372. \]

7. **Combine the terms:**
   \[ f'(x) = \frac{4x^3 + 56x^2 + 196x - 4x^3 - 28x^2 - 196x - 1372}{(x^2 + 14x + 49)^2}, \]
   \[ f'(x) = \frac{28x^2 - 1372}{(x^2 + 14x + 49)^2}. \]

8. **Set the derivative equal to zero to find critical points:**
   \[ 28x^2 - 1372 = 0, \]
   \[ x^2 = \frac{1372}{28}, \]
   \[ x^2 = 49, \]
   \[ x = \pm 7. \]

9. **Evaluate the function at the critical points:**
   Since \( x = -7 \) is not in the domain, we only consider \( x = 7 \):
   \[ f(7) = \frac{2(7)^2 + 98}{(7 + 7)^2} = \frac{2 \cdot 49 + 98}{14^2} = \frac{98 + 98}{196} = \frac{196}{196} = 1. \]

10. **Check the behavior as \( x \to \infty \) and \( x \to -\infty \):**
    \[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{2x^2 + 98}{x^2 + 14x + 49} = 2. \]
    \[ \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{2x^2 + 98}{x^2 + 14x + 49} = 2. \]

11. **Conclusion:**
    The smallest value of the expression is achieved at \( x = 7 \), and the value is \( 1 \).

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