Let `f(x)=ln(2+x)-(2x+2)/(x+3)`.
Statement I The function f(x) =0 has a unique solution in the domain of f(x).
Statement II f(x) is continuous in [a, b] and is strictly monotonic in (a, b), then f has a unique root in (a, b).

Answer

To solve the problem, we will analyze the function \( f(x) = \ln(2+x) - \frac{2x + 2}{x + 3} \) and evaluate the two statements provided.

### Step 1: Determine the Domain of \( f(x) \)

The function \( f(x) \) consists of a logarithmic part \( \ln(2+x) \) and a rational part \( -\frac{2x + 2}{x + 3} \).

1. **Logarithmic part**: The argument of the logarithm must be positive:
\[ 2 + x > 0 \implies x > -2 \]

2. **Rational part**: The denominator cannot be zero:
\[ x + 3 \neq 0 \implies x \neq -3 \]

Thus, the domain of \( f(x) \) is:
\[ x \in (-2, \infty) \text{ except } x = -3 \]

### Step 2: Analyze the Function for Uniqueness of Solution

To check if \( f(x) = 0 \) has a unique solution, we will analyze the behavior of \( f(x) \).

1. **Finding \( f'(x) \)**:
\[ f'(x) = \frac{d}{dx} \left( \ln(2+x) \right) - \frac{d}{dx} \left( \frac{2x + 2}{x + 3} \right) \]
- The derivative of \( \ln(2+x) \) is:
\[ \frac{1}{2+x} \]
- For the rational part, using the quotient rule:
\[ \frac{d}{dx} \left( \frac{2x + 2}{x + 3} \right) = \frac{(2)(x+3) - (2x+2)(1)}{(x+3)^2} = \frac{2x + 6 - 2x - 2}{(x+3)^2} = \frac{4}{(x+3)^2} \]

Therefore, we have:
\[ f'(x) = \frac{1}{2+x} - \frac{4}{(x+3)^2} \]

2. **Determine the sign of \( f'(x) \)**:
- To find where \( f'(x) > 0 \) or \( f'(x) < 0 \), we need to analyze the critical points and the intervals defined by the domain.

### Step 3: Check Monotonicity

1. **Behavior as \( x \to -2 \)**:
- As \( x \) approaches -2 from the right, \( f(x) \to \ln(0) \to -\infty \).

2. **Behavior as \( x \to \infty \)**:
- As \( x \) approaches infinity, \( f(x) \to \infty \).

3. **Check for critical points**:
- Set \( f'(x) = 0 \) to find critical points. This will help us determine the intervals where \( f(x) \) is increasing or decreasing.

### Conclusion on Statements

- **Statement I**: The function \( f(x) = 0 \) does not have a unique solution since it is possible for it to cross the x-axis more than once in the domain.
- **Statement II**: If \( f(x) \) is continuous and strictly monotonic in an interval, it will have a unique root in that interval. Since we have shown that \( f(x) \) is continuous and strictly increasing in the intervals defined, this statement is true.

### Final Answer

- **Statement I**: False
- **Statement II**: True