Solve the inequality `(x-1)/(x^(2)-4x+3) lt 1 `

Answer

To solve the inequality \(\frac{x-1}{x^2-4x+3} < 1\), we will follow these steps:

### Step 1: Rearrange the Inequality
We start by moving 1 to the left side of the inequality:
\[ \frac{x-1}{x^2-4x+3} - 1 < 0 \]

### Step 2: Find a Common Denominator
To combine the terms, we need a common denominator, which is \(x^2 - 4x + 3\):
\[ \frac{x-1 - (x^2 - 4x + 3)}{x^2 - 4x + 3} < 0 \]

### Step 3: Simplify the Numerator
Now, simplify the numerator:
\[ x - 1 - (x^2 - 4x + 3) = x - 1 - x^2 + 4x - 3 = -x^2 + 5x - 4 \]
Thus, we rewrite the inequality as:
\[ \frac{-x^2 + 5x - 4}{x^2 - 4x + 3} < 0 \]

### Step 4: Factor the Expressions
Next, we factor both the numerator and the denominator:
- The numerator \(-x^2 + 5x - 4\) can be factored as \(-(x-4)(x-1)\).
- The denominator \(x^2 - 4x + 3\) factors to \((x-3)(x-1)\).

Putting it all together, we have:
\[ \frac{-(x-4)(x-1)}{(x-3)(x-1)} < 0 \]

### Step 5: Cancel Common Factors
We can cancel the common factor \((x-1)\) from the numerator and denominator, but we must note that \(x \neq 1\):
\[ \frac{-(x-4)}{(x-3)} < 0 \]

### Step 6: Analyze the Inequality
Now we analyze the inequality:
\[ \frac{-(x-4)}{(x-3)} < 0 \]
This inequality will be negative when the numerator and denominator have opposite signs.

### Step 7: Determine Critical Points
The critical points are \(x = 4\) and \(x = 3\). We also need to consider \(x = 1\) where the original expression is undefined.

### Step 8: Test Intervals
We will test the intervals determined by these critical points:
1. \( (-\infty, 1) \)
2. \( (1, 3) \)
3. \( (3, 4) \)
4. \( (4, \infty) \)

- For \(x < 1\) (e.g., \(x = 0\)):
\[ \frac{-(0-4)}{(0-3)} = \frac{4}{-3} < 0 \quad \text{(True)} \]

- For \(1 < x < 3\) (e.g., \(x = 2\)):
\[ \frac{-(2-4)}{(2-3)} = \frac{2}{-1} < 0 \quad \text{(True)} \]

- For \(3 < x < 4\) (e.g., \(x = 3.5\)):
\[ \frac{-(3.5-4)}{(3.5-3)} = \frac{0.5}{0.5} > 0 \quad \text{(False)} \]

- For \(x > 4\) (e.g., \(x = 5\)):
\[ \frac{-(5-4)}{(5-3)} = \frac{-1}{2} < 0 \quad \text{(True)} \]

### Step 9: Combine the Results
From our tests, the inequality is satisfied in the intervals:
\[ (-\infty, 1) \cup (1, 3) \cup (4, \infty) \]

### Final Solution
Thus, the solution to the inequality \(\frac{x-1}{x^2-4x+3} < 1\) is:
\[ x \in (-\infty, 1) \cup (1, 3) \cup (4, \infty) \]