Solve : `(i)" "((x-1)\(x-2)(x-3))/((x+1)(x+2)(x+3))" "(ii) " "(x^(4)+x^(2)+1)/(x^(2)+4x-5)lt0`

Answer

To solve the given problems step by step, we will address each part of the question separately.

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

**Step 1: Identify the critical points.**
The critical points occur when the numerator or denominator equals zero.
- Numerator: \( (x-1)(x-2)(x-3) = 0 \) gives \( x = 1, 2, 3 \).
- Denominator: \( (x+1)(x+2)(x+3) = 0 \) gives \( x = -1, -2, -3 \).

**Step 2: Determine the intervals.**
The critical points divide the number line into intervals:
- \( (-\infty, -3) \)
- \( (-3, -2) \)
- \( (-2, -1) \)
- \( (-1, 1) \)
- \( (1, 2) \)
- \( (2, 3) \)
- \( (3, \infty) \)

**Step 3: Test each interval.**
Choose a test point from each interval to determine the sign of the expression.

1. For \( x = -4 \) in \( (-\infty, -3) \):
\[ \frac{(-)(-)(-)}{(-)(-)(-)} = \frac{-}{-} = + \]

2. For \( x = -2.5 \) in \( (-3, -2) \):
\[ \frac{(-)(-)(-)}{(-)(+)(-)} = \frac{-}{+} = - \]

3. For \( x = -1.5 \) in \( (-2, -1) \):
\[ \frac{(-)(-)(-)}{(+)(+)(-)} = \frac{-}{-} = + \]

4. For \( x = 0 \) in \( (-1, 1) \):
\[ \frac{(-)(-)(-)}{(+)(+)(+)} = \frac{-}{+} = - \]

5. For \( x = 1.5 \) in \( (1, 2) \):
\[ \frac{(+)(-)(-)}{(+)(+)(+)} = \frac{+}{+} = + \]

6. For \( x = 2.5 \) in \( (2, 3) \):
\[ \frac{(+)(+)(-)}{(+)(+)(+)} = \frac{-}{+} = - \]

7. For \( x = 4 \) in \( (3, \infty) \):
\[ \frac{(+)(+)(+)}{(+)(+)(+)} = \frac{+}{+} = + \]

**Step 4: Compile the results.**
The expression is positive in the intervals:
- \( (-\infty, -3) \)
- \( (-2, -1) \)
- \( (1, 2) \)
- \( (3, \infty) \)

**Step 5: Write the final solution.**
The solution to the inequality is:
\[ x \in (-\infty, -3) \cup (-2, -1) \cup (1, 2) \cup (3, \infty) \]

### Part (ii): Solve \(\frac{x^4 + x^2 + 1}{x^2 - 4x - 5} < 0\)

**Step 1: Analyze the numerator.**
The numerator \( x^4 + x^2 + 1 \) is always positive because it is a sum of squares (even degree polynomial with positive leading coefficient).

**Step 2: Factor the denominator.**
Factor \( x^2 - 4x - 5 \):
\[ x^2 - 4x - 5 = (x - 5)(x + 1) \]

**Step 3: Identify the critical points.**
The critical points from the denominator are:
- \( x = 5 \)
- \( x = -1 \)

**Step 4: Determine the intervals.**
The critical points divide the number line into intervals:
- \( (-\infty, -1) \)
- \( (-1, 5) \)
- \( (5, \infty) \)

**Step 5: Test each interval.**
Choose a test point from each interval to determine the sign of the expression.

1. For \( x = -2 \) in \( (-\infty, -1) \):
\[ \frac{(+)}{(-)} = - \]

2. For \( x = 0 \) in \( (-1, 5) \):
\[ \frac{(+)}{(-)} = - \]

3. For \( x = 6 \) in \( (5, \infty) \):
\[ \frac{(+)}{(+)} = + \]

**Step 6: Compile the results.**
The expression is negative in the intervals:
- \( (-\infty, -1) \)
- \( (-1, 5) \)

**Step 7: Write the final solution.**
The solution to the inequality is:
\[ x \in (-\infty, -1) \cup (-1, 5) \]