Solve : `(x+4)/(x-2)-(2)/(x+1) lt 0`

Answer

To solve the inequality \(\frac{x+4}{x-2} - \frac{2}{x+1} < 0\), we will follow these steps:

### Step 1: Find a common denominator
The common denominator for the fractions \(\frac{x+4}{x-2}\) and \(\frac{2}{x+1}\) is \((x-2)(x+1)\).

### Step 2: Rewrite the inequality
We can rewrite the inequality as:
\[ \frac{(x+4)(x+1) - 2(x-2)}{(x-2)(x+1)} < 0 \]

### Step 3: Expand the numerator
Now, we will expand the numerator:
\[ (x+4)(x+1) = x^2 + 5x + 4 \]
\[ 2(x-2) = 2x - 4 \]
So, the complete numerator becomes:
\[ x^2 + 5x + 4 - (2x - 4) = x^2 + 5x + 4 - 2x + 4 = x^2 + 3x + 8 \]

### Step 4: Rewrite the inequality
Now we have:
\[ \frac{x^2 + 3x + 8}{(x-2)(x+1)} < 0 \]

### Step 5: Analyze the numerator
Next, we need to analyze the numerator \(x^2 + 3x + 8\). To determine if it can be negative, we can check its discriminant:
\[ D = b^2 - 4ac = 3^2 - 4(1)(8) = 9 - 32 = -23 \]
Since the discriminant is negative, the quadratic \(x^2 + 3x + 8\) is always positive.

### Step 6: Analyze the denominator
Now we need to analyze the denominator \((x-2)(x+1)\). We want to find when this expression is less than zero:
\[ (x-2)(x+1) < 0 \]

### Step 7: Find the critical points
The critical points are \(x = 2\) and \(x = -1\). We will test the intervals determined by these points:
- Interval 1: \( (-\infty, -1) \)
- Interval 2: \( (-1, 2) \)
- Interval 3: \( (2, \infty) \)

### Step 8: Test the intervals
1. For \(x < -1\) (e.g., \(x = -2\)):
\((x-2)(x+1) = (-2-2)(-2+1) = (-4)(-1) > 0\)

2. For \(-1 < x < 2\) (e.g., \(x = 0\)):
\((x-2)(x+1) = (0-2)(0+1) = (-2)(1) < 0\)

3. For \(x > 2\) (e.g., \(x = 3\)):
\((x-2)(x+1) = (3-2)(3+1) = (1)(4) > 0\)

### Step 9: Conclusion
The expression \(\frac{x^2 + 3x + 8}{(x-2)(x+1)} < 0\) is satisfied only in the interval \((-1, 2)\).

Thus, the solution to the inequality is:
\[ \boxed{(-1, 2)} \]