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Let's solve the given inequalities step by step.

### 1. \((-x + 3)(1 - x) < 0\)
We need to find when the product of these two factors is negative. Start by determining the roots of each factor:
\[
-x + 3 = 0 \quad \Rightarrow \quad x = 3
\]
\[
1 - x = 0 \quad \Rightarrow \quad x = 1
\]

These two points, \(x = 1\) and \(x = 3\), divide the number line into three intervals: \((- \infty, 1)\), \((1, 3)\), and \((3, \infty)\).

Now, we test the sign of the product in each interval:
- For \(x < 1\), say \(x = 0\): \((-x + 3)(1 - x) = (3)(1) > 0\)
- For \(1 < x < 3\), say \(x = 2\): \((-x + 3)(1 - x) = (1)(-1) < 0\)
- For \(x > 3\), say \(x = 4\): \((-x + 3)(1 - x) = (-1)(-3) > 0\)

Therefore, the inequality holds in the interval \((1, 3)\).

### Solution for 1: \( x \in (1, 3) \).

---

### 2. \(25x^2 - 9 \geq 0\)
This is a quadratic inequality. First, rewrite the expression:
\[
25x^2 - 9 = (5x - 3)(5x + 3)
\]
We now solve the corresponding equation:
\[
(5x - 3)(5x + 3) = 0 \quad \Rightarrow \quad x = \pm \frac{3}{5}
\]
This divides the number line into three intervals: \((- \infty, -\frac{3}{5})\), \((- \frac{3}{5}, \frac{3}{5})\), and \((\frac{3}{5}, \infty)\).

Now, test the sign of the product in each interval:
- For \(x < -\frac{3}{5}\), say \(x = -1\): \((5x - 3)(5x + 3) = (-8)(-2) > 0\)
- For \(-\frac{3}{5} < x < \frac{3}{5}\), say \(x = 0\): \((5x - 3)(5x + 3) = (-3)(3) < 0\)
- For \(x > \frac{3}{5}\), say \(x = 1\): \((5x - 3)(5x + 3) = (2)(8) > 0\)

Since we are looking for the intervals where the expression is \(\geq 0\), we include the points where it equals zero and the positive intervals.

### Solution for 2: \(x \in (-\infty, -\frac{3}{5}] \cup [\frac{3}{5}, \infty)\).

---

### 3. \((2x + 5)(-2x + 7) \leq 0\)
We begin by finding the roots of each factor:
\[
2x + 5 = 0 \quad \Rightarrow \quad x = -\frac{5}{2}
\]
\[
-2x + 7 = 0 \quad \Rightarrow \quad x = \frac{7}{2}
\]

These two points divide the number line into three intervals: \((- \infty, -\frac{5}{2})\), \((- \frac{5}{2}, \frac{7}{2})\), and \((\frac{7}{2}, \infty)\).

Now, we test the sign of the product in each interval:
- For \(x < -\frac{5}{2}\), say \(x = -3\): \((2x + 5)(-2x + 7) = (-1)(13) < 0\)
- For \(-\frac{5}{2} < x < \frac{7}{2}\), say \(x = 0\): \((2x + 5)(-2x + 7) = (5)(7) > 0\)
- For \(x > \frac{7}{2}\), say \(x = 4\): \((2x + 5)(-2x + 7) = (13)(-1) < 0\)

Since we want the product to be less than or equal to zero, we include the points where it equals zero.

### Solution for 3: \(x \in (-\infty, -\frac{5}{2}] \cup [\frac{7}{2}, \infty)\).

---

Would you like further details or explanations on any of these solutions?  

### Related questions:
1. How do you solve a quadratic inequality in general?
2. Can you explain the steps for solving inequalities with products in more detail?
3. How do you determine the critical points for inequality solutions?
4. Why do we test intervals when solving inequalities?
5. How would the solutions change if the inequalities were strict inequalities?

### Tip:
When solving inequalities involving products or quotients, always check the critical points and test the intervals between them to determine where the inequality holds.

Résoudre dans ℝ les inéquations suivantes: 1. (-x + 3)(1 - x) < 0, 2. 25x^2 - 9 ≥ 0, 3. (2x + 5)(-2x + 7) ≤ 0.

Explore how to solve inequalities involving polynomial products and quadratic equations, such as (-x + 3)(1 - x) < 0 and 25x^2 - 9 ≥ 0. The solution involves factoring, sign analysis, and solving quadratic equations. Suitable for students in Grades 10-12.

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