Solve the following inequalities:
`((x-3)(x+2))/(x^(2)-1)lt1`

Answer

To solve the inequality \(\frac{(x-3)(x+2)}{x^2-1} < 1\), we will follow these steps:

### Step 1: Rearrange the Inequality
We start by moving 1 to the left side of the inequality:
\[ \frac{(x-3)(x+2)}{x^2-1} - 1 < 0 \]
This can be rewritten as:
\[ \frac{(x-3)(x+2) - (x^2-1)}{x^2-1} < 0 \]

### Step 2: Simplify the Numerator
Now, we simplify the numerator:
\[ (x-3)(x+2) - (x^2 - 1) = (x^2 - x - 6) - (x^2 - 1) \]
This simplifies to:
\[ -x - 5 \]
So, we can rewrite the inequality as:
\[ \frac{-x - 5}{x^2 - 1} < 0 \]

### Step 3: Factor the Denominator
Next, we factor the denominator:
\[ x^2 - 1 = (x-1)(x+1) \]
Thus, the inequality becomes:
\[ \frac{-x - 5}{(x-1)(x+1)} < 0 \]

### Step 4: Find Critical Points
Now we find the critical points by setting the numerator and denominator to zero:
1. **Numerator**: \(-x - 5 = 0 \Rightarrow x = -5\)
2. **Denominator**: \((x-1)(x+1) = 0 \Rightarrow x = 1, -1\)

The critical points are \(x = -5, -1, 1\).

### Step 5: Test Intervals
We will test the sign of the expression in the intervals determined by these critical points:
- \((-∞, -5)\)
- \((-5, -1)\)
- \((-1, 1)\)
- \((1, ∞)\)

1. **Interval \((-∞, -5)\)**: Choose \(x = -6\)
\[ \frac{-(-6) - 5}{(-6-1)(-6+1)} = \frac{6 - 5}{(-7)(-5)} = \frac{1}{35} > 0 \]

2. **Interval \((-5, -1)\)**: Choose \(x = -3\)
\[ \frac{-(-3) - 5}{(-3-1)(-3+1)} = \frac{3 - 5}{(-4)(-2)} = \frac{-2}{8} < 0 \]

3. **Interval \((-1, 1)\)**: Choose \(x = 0\)
\[ \frac{-(0) - 5}{(0-1)(0+1)} = \frac{-5}{(-1)(1)} = 5 > 0 \]

4. **Interval \((1, ∞)\)**: Choose \(x = 2\)
\[ \frac{-(2) - 5}{(2-1)(2+1)} = \frac{-2 - 5}{(1)(3)} = \frac{-7}{3} < 0 \]

### Step 6: Determine the Solution Set
From our tests, we find that the expression is negative in the intervals:
- \((-5, -1)\)
- \((1, ∞)\)

### Step 7: Write the Final Solution
Thus, the solution to the inequality \(\frac{(x-3)(x+2)}{x^2-1} < 1\) is:
\[ x \in (-5, -1) \cup (1, \infty) \]