Determine the values of x satisfying the following equalities:
(a) `|(x-1)/(x+1)|=(x-1)/(x+1)` (b) `|x^(2)-5x+6|=-(x^(2)+-5x+6)`

Answer

To solve the given equalities, we will break them down step by step.

### Part (a): Solve `|(x-1)/(x+1)| = (x-1)/(x+1)`

1. **Understanding the Absolute Value**:
The equation `|(x-1)/(x+1)| = (x-1)/(x+1)` implies that the expression `(x-1)/(x+1)` is either non-negative or zero. This leads us to two cases:
- Case 1: `(x-1)/(x+1) >= 0`
- Case 2: `(x-1)/(x+1) < 0` (which we will not consider since it contradicts the equality)

2. **Finding Critical Points**:
The critical points occur when the numerator or denominator is zero:
- Numerator: `x - 1 = 0` → `x = 1`
- Denominator: `x + 1 = 0` → `x = -1`

3. **Test Intervals**:
We will test the sign of `(x-1)/(x+1)` in the intervals determined by the critical points:
- Interval 1: `(-∞, -1)`
- Interval 2: `(-1, 1)`
- Interval 3: `(1, ∞)`

- For `x < -1`: Choose `x = -2`, `(x-1)/(x+1) = (-2-1)/(-2+1) = -3/-1 = 3` (positive)
- For `-1 < x < 1`: Choose `x = 0`, `(x-1)/(x+1) = (0-1)/(0+1) = -1` (negative)
- For `x > 1`: Choose `x = 2`, `(x-1)/(x+1) = (2-1)/(2+1) = 1/3` (positive)

4. **Conclusion from Cases**:
From the tests, we find:
- `(x-1)/(x+1) >= 0` for `x ∈ (-∞, -1) ∪ (1, ∞)`

5. **Final Solution**:
The values of `x` that satisfy the equation are:
\[ x \in (-\infty, -1) \cup (1, \infty) \]

### Part (b): Solve `|x^2 - 5x + 6| = -(x^2 - 5x + 6)`

1. **Understanding the Absolute Value**:
The equation `|x^2 - 5x + 6| = -(x^2 - 5x + 6)` implies that the expression `x^2 - 5x + 6` must be negative or zero. This leads us to:
- Case 1: `x^2 - 5x + 6 < 0`
- Case 2: `x^2 - 5x + 6 = 0`

2. **Finding Roots**:
We can factor the quadratic:
\[ x^2 - 5x + 6 = (x-2)(x-3) \]
The roots are `x = 2` and `x = 3`.

3. **Test Intervals**:
We will test the sign of `x^2 - 5x + 6` in the intervals:
- Interval 1: `(-∞, 2)`
- Interval 2: `(2, 3)`
- Interval 3: `(3, ∞)`

- For `x < 2`: Choose `x = 0`, `x^2 - 5x + 6 = 6` (positive)
- For `2 < x < 3`: Choose `x = 2.5`, `x^2 - 5x + 6 = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25` (negative)
- For `x > 3`: Choose `x = 4`, `x^2 - 5x + 6 = 16 - 20 + 6 = 2` (positive)

4. **Conclusion from Cases**:
The expression `x^2 - 5x + 6 < 0` holds for `x ∈ (2, 3)`.

5. **Final Solution**:
The values of `x` that satisfy the equation are:
\[ x \in [2, 3] \]

### Summary of Solutions:
- Part (a): \( x \in (-\infty, -1) \cup (1, \infty) \)
- Part (b): \( x \in [2, 3] \)