To solve the given inequalities step by step, we will break down each part of the question.
### Part (i): Solve the inequality
\[
\frac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)} < 0
\]
**Step 1: Identify the domain of the function**
- The function is undefined when the denominator is zero. Therefore, we need to find the values of \(x\) that make the denominator zero:
- \(x + 1 = 0 \Rightarrow x = -1\)
- \(x + 2 = 0 \Rightarrow x = -2\)
- \(x + 3 = 0 \Rightarrow x = -3\)
Thus, the function is undefined at \(x = -1, -2, -3\).
**Hint:** Check where the denominator equals zero to find points of discontinuity.
**Step 2: Find the critical points**
- The critical points from the numerator are:
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(x - 2 = 0 \Rightarrow x = 2\)
- \(x - 3 = 0 \Rightarrow x = 3\)
So, the critical points are \(x = -3, -2, -1, 1, 2, 3\).
**Hint:** Identify both the zeros of the numerator and the denominator.
**Step 3: Create a number line and test 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 4: Test each interval**
- Choose test points from each interval to determine the sign of the expression:
- For \(x < -3\) (e.g., \(x = -4\)): \(\frac{(-)(-)(-)}{(-)(-)(-)} = -\) (negative)
- For \((-3, -2)\) (e.g., \(x = -2.5\)): \(\frac{(-)(-)(-)}{(+)(-)(-)} = +\) (positive)
- For \((-2, -1)\) (e.g., \(x = -1.5\)): \(\frac{(-)(-)(-)}{(+)(+)(-)} = -\) (negative)
- For \((-1, 1)\) (e.g., \(x = 0\)): \(\frac{(-)(-)(-)}{(+)(+)(+)} = -\) (negative)
- For \((1, 2)\) (e.g., \(x = 1.5\)): \(\frac{(+)(-)(-)}{(+)(+)(+)} = +\) (positive)
- For \((2, 3)\) (e.g., \(x = 2.5\)): \(\frac{(+)(+)(-)}{(+)(+)(+)} = -\) (negative)
- For \(x > 3\) (e.g., \(x = 4\)): \(\frac{(+)(+)(+)}{(+)(+)(+)} = +\) (positive)
**Step 5: Determine where the expression is negative**
- The intervals where the expression is negative are:
- \((- \infty, -3)\)
- \((-2, -1)\)
- \((-1, 1)\)
- \((2, 3)\)
**Step 6: Write the final solution**
- The solution to the inequality is:
\[
x \in (-\infty, -3) \cup (-2, -1) \cup (-1, 1) \cup (2, 3)
\]
### Part (ii): Solve the inequality
\[
\frac{x^4 + x^2 + 1}{x^2 + 4x - 5} < 0
\]
**Step 1: Analyze the numerator**
- The numerator \(x^4 + x^2 + 1\) is always positive because:
- \(x^4\) is non-negative for all \(x\)
- \(x^2\) is non-negative for all \(x\)
- The constant \(1\) is positive.
**Hint:** Check the behavior of the numerator to determine if it can be negative.
**Step 2: Analyze the denominator**
- Set the denominator equal to zero to find critical points:
\[
x^2 + 4x - 5 = 0
\]
Factoring gives:
\[
(x + 5)(x - 1) = 0 \Rightarrow x = -5, 1
\]
**Step 3: Determine the domain**
- The function is undefined at \(x = -5\) and \(x = 1\). Thus, the domain is:
\[
x \in (-\infty, -5) \cup (-5, 1) \cup (1, \infty)
\]
**Step 4: Test the intervals**
- The critical points divide the number line into intervals:
- \((- \infty, -5)\)
- \((-5, 1)\)
- \((1, \infty)\)
- Choose test points:
- For \(x < -5\) (e.g., \(x = -6\)): \(\text{positive} / \text{negative} = -\) (negative)
- For \((-5, 1)\) (e.g., \(x = 0\)): \(\text{positive} / \text{positive} = +\) (positive)
- For \(x > 1\) (e.g., \(x = 2\)): \(\text{positive} / \text{positive} = +\) (positive)
**Step 5: Determine where the expression is negative**
- The only interval where the expression is negative is:
\((- \infty, -5)\)
**Step 6: Write the final solution**
- The solution to the inequality is:
\[
x \in (-\infty, -5)
\]
### Summary of Solutions
1. For Part (i):
\[
x \in (-\infty, -3) \cup (-2, -1) \cup (-1, 1) \cup (2, 3)
\]
2. For Part (ii):
\[
x \in (-\infty, -5)
\]
