The solution set of the inequation `(|x+3|+x)/(x+2) gt 1`, is

Options

A. `(-5, -2) cup (-1, oo)`
B. `(-5, -2)`
C. `(-1, oo)`
D. none of these

Explanation

To solve the inequality \(\frac{|x+3| + x}{x+2} > 1\), we will break it down into manageable steps.

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \frac{|x+3| + x}{x+2} > 1 \]
Subtract 1 from both sides:
\[ \frac{|x+3| + x}{x+2} - 1 > 0 \]
This can be rewritten as:
\[ \frac{|x+3| + x - (x + 2)}{x+2} > 0 \]
Simplifying the numerator gives:
\[ \frac{|x+3| - 2}{x+2} > 0 \]

### Step 2: Analyze the Absolute Value
Next, we consider two cases based on the expression inside the absolute value, \(x + 3\).

**Case 1:** \(x + 3 \geq 0 \Rightarrow x \geq -3\)

In this case, \(|x + 3| = x + 3\). Substitute this into the inequality:
\[ \frac{(x + 3) - 2}{x + 2} > 0 \]
This simplifies to:
\[ \frac{x + 1}{x + 2} > 0 \]

**Case 2:** \(x + 3 < 0 \Rightarrow x < -3\)

Here, \(|x + 3| = -(x + 3) = -x - 3\). Substitute this into the inequality:
\[ \frac{(-x - 3) - 2}{x + 2} > 0 \]
This simplifies to:
\[ \frac{-x - 5}{x + 2} > 0 \]
We can multiply both sides by -1 (which flips the inequality):
\[ \frac{x + 5}{x + 2} < 0 \]

### Step 3: Solve Each Case
**For Case 1:** \(\frac{x + 1}{x + 2} > 0\)

The critical points are \(x = -1\) and \(x = -2\). We test intervals:
- For \(x < -2\): Both numerator and denominator are negative, so the fraction is positive.
- For \(-2 < x < -1\): The numerator is negative and the denominator is positive, so the fraction is negative.
- For \(x > -1\): Both numerator and denominator are positive, so the fraction is positive.

Thus, the solution for Case 1 is:
\[ x \in (-\infty, -2) \cup (-1, \infty) \]
However, we must also satisfy \(x \geq -3\), so we restrict this to:
\[ x \in [-3, -2) \cup (-1, \infty) \]

**For Case 2:** \(\frac{x + 5}{x + 2} < 0\)

The critical points are \(x = -5\) and \(x = -2\). We test intervals:
- For \(x < -5\): Both numerator and denominator are negative, so the fraction is positive.
- For \(-5 < x < -2\): The numerator is positive and the denominator is negative, so the fraction is negative.
- For \(x > -2\): Both numerator and denominator are positive, so the fraction is positive.

Thus, the solution for Case 2 is:
\[ x \in (-5, -2) \]
And since we need \(x < -3\), we restrict this to:
\[ x \in (-5, -3) \]

### Step 4: Combine the Solutions
Now, we combine the solutions from both cases:
1. From Case 1: \(x \in [-3, -2) \cup (-1, \infty)\)
2. From Case 2: \(x \in (-5, -3)\)

Combining these gives:
\[ x \in (-5, -2) \cup (-1, \infty) \]

### Final Answer
The solution set of the inequation \(\frac{|x+3| + x}{x+2} > 1\) is:
\[ \boxed{(-5, -2) \cup (-1, \infty)} \]