Solve the following inequalities.
`(x^(2)-2x+1)/(x+1)>0`

Answer

To solve the inequality \(\frac{x^2 - 2x + 1}{x + 1} > 0\), we will follow these steps:

### Step 1: Factor the numerator
The numerator \(x^2 - 2x + 1\) can be factored as:
\[ x^2 - 2x + 1 = (x - 1)^2 \]
So, we can rewrite the inequality as:
\[ \frac{(x - 1)^2}{x + 1} > 0 \]

### Step 2: Analyze the factors
Next, we need to analyze the expression \((x - 1)^2\) and \(x + 1\):
- The term \((x - 1)^2\) is always non-negative (i.e., \((x - 1)^2 \geq 0\)) because it is a square.
- The term \(x + 1\) changes sign depending on the value of \(x\).

### Step 3: Determine when the expression is positive
The expression \(\frac{(x - 1)^2}{x + 1}\) is positive when:
1. The numerator \((x - 1)^2 > 0\) (which is true for all \(x \neq 1\))
2. The denominator \(x + 1 > 0\) (which is true for \(x > -1\))

### Step 4: Find critical points
We need to find the critical points where the expression could be zero or undefined:
- The numerator is zero at \(x = 1\).
- The denominator is zero at \(x = -1\).

### Step 5: Test intervals
We will test the sign of the expression in the intervals determined by the critical points \(x = -1\) and \(x = 1\):
1. **Interval \((-∞, -1)\)**: Choose \(x = -2\):
\[ \frac{(-2 - 1)^2}{-2 + 1} = \frac{9}{-1} < 0 \quad \text{(negative)} \]
2. **Interval \((-1, 1)\)**: Choose \(x = 0\):
\[ \frac{(0 - 1)^2}{0 + 1} = \frac{1}{1} > 0 \quad \text{(positive)} \]
3. **Interval \((1, ∞)\)**: Choose \(x = 2\):
\[ \frac{(2 - 1)^2}{2 + 1} = \frac{1}{3} > 0 \quad \text{(positive)} \]

### Step 6: Combine results
From our tests, we find:
- The expression is negative in \((-∞, -1)\).
- The expression is positive in \((-1, 1)\) and \((1, ∞)\).

### Step 7: Write the solution
Since the inequality is strict (\(> 0\)), we do not include the points where the expression is zero or undefined:
\[ \text{Solution: } (-1, 1) \cup (1, \infty) \]