Solve the following rational in equalities
`(i) ((x-1)(x+2))/((x-3)(x+3)) lt 0` `(ii) ((1-x)^(3)(x+2)^(4))/((x+9)^(2)(x-8))ge0`
`(iii) ((x^(2)-3x+1)^(3))/((x-1)(x+2))le0` `(iv) (x(2^(x)-3^(x)))/((x^(2)+x+1)(x-1))gt0`
`(v) ((x-1)(x-2)(x-3))/((x+1)(x+2)(x+3)) le1` `(vi) 1 lt (3x^(2)-7x+8)/(x^(2)+1) le2`

Answer

To solve the given rational inequalities step by step, we'll follow a systematic approach for each part:

### (i) Solve \(\frac{(x-1)(x+2)}{(x-3)(x+3)} < 0\)

1. **Identify critical points**: Set the numerator and denominator to zero.
- \(x-1 = 0 \Rightarrow x = 1\)
- \(x+2 = 0 \Rightarrow x = -2\)
- \(x-3 = 0 \Rightarrow x = 3\)
- \(x+3 = 0 \Rightarrow x = -3\)

2. **Critical points**: The critical points are \(x = -3, -2, 1, 3\).

3. **Number line**: Place the critical points on a number line and test intervals:
- Intervals: \((-∞, -3)\), \((-3, -2)\), \((-2, 1)\), \((1, 3)\), \((3, ∞)\)

4. **Test each interval**:
- For \(x < -3\) (e.g., \(x = -4\)): \(\frac{(-)(-)}{(-)(-)} > 0\)
- For \(-3 < x < -2\) (e.g., \(x = -2.5\)): \(\frac{(-)(+)}{(-)(-)} < 0\)
- For \(-2 < x < 1\) (e.g., \(x = 0\)): \(\frac{(-)(+)}{(-)(+)} > 0\)
- For \(1 < x < 3\) (e.g., \(x = 2\)): \(\frac{(+)(+)}{(-)(+)} < 0\)
- For \(x > 3\) (e.g., \(x = 4\)): \(\frac{(+)(+)}{(+)(+)} > 0\)

5. **Solution**: The intervals where the expression is negative are \((-3, -2)\) and \((1, 3)\).

**Final answer**: \(x \in (-3, -2) \cup (1, 3)\)