https://file.aiquickdraw.com/mathbot/file/1727894594090vl0kizwd.jpg

The image you uploaded contains several exercises. I will start by solving **Exercise 3** as it is clear from the photo.

### Exercise 3

#### 1) Développer et réduire ce qui suit:
\[
o = \sqrt{7}(5 - 2\sqrt{7}) - (5 - 2\sqrt{7})(5 + 2\sqrt{7}) + (5 + 2\sqrt{7})^2
\]

Let's break this expression down step by step.

- First term: 
  \[
  \sqrt{7}(5 - 2\sqrt{7}) = 5\sqrt{7} - 2\sqrt{7} \cdot \sqrt{7} = 5\sqrt{7} - 2 \cdot 7 = 5\sqrt{7} - 14
  \]

- Second term:
  The second term is the product of two conjugates:
  \[
  (5 - 2\sqrt{7})(5 + 2\sqrt{7}) = 5^2 - (2\sqrt{7})^2 = 25 - 4 \cdot 7 = 25 - 28 = -3
  \]

- Third term:
  The third term is the square of a binomial:
  \[
  (5 + 2\sqrt{7})^2 = 5^2 + 2 \cdot 5 \cdot 2\sqrt{7} + (2\sqrt{7})^2 = 25 + 20\sqrt{7} + 4 \cdot 7 = 25 + 20\sqrt{7} + 28 = 53 + 20\sqrt{7}
  \]

Now, let's combine everything:
\[
5\sqrt{7} - 14 - (-3) + (53 + 20\sqrt{7})
\]
\[
= 5\sqrt{7} - 14 + 3 + 53 + 20\sqrt{7}
\]
\[
= (5\sqrt{7} + 20\sqrt{7}) + (-14 + 3 + 53)
\]
\[
= 25\sqrt{7} + 42
\]

So, the simplified result is:
\[
o = 25\sqrt{7} + 42
\]

#### 2) Résoudre les équations suivantes:

a) \( 5x^2 - 3 = 12 \)

Step 1: Add 3 to both sides.
\[
5x^2 = 15
\]

Step 2: Divide by 5.
\[
x^2 = 3
\]

Step 3: Solve for \(x\).
\[
x = \pm \sqrt{3}
\]

So, the solutions are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).

b) \( 2 - x^2 - 6x = 3(1 - 2x) - 5 \)

Expand and simplify both sides:

Left-hand side:
\[
2 - x^2 - 6x
\]

Right-hand side:
\[
3(1 - 2x) - 5 = 3 - 6x - 5 = -2 - 6x
\]

Now, equate both sides:
\[
2 - x^2 - 6x = -2 - 6x
\]

Simplify:
\[
2 - x^2 = -2
\]

Move all terms to one side:
\[
x^2 = 4
\]

Solve for \(x\):
\[
x = \pm 2
\]

So, the solutions are \(x = 2\) and \(x = -2\).

c) \( (2x + 1)(3 - x) = 5x(1 + 4x) \)

Expand both sides:

Left-hand side:
\[
(2x + 1)(3 - x) = 6x - 2x^2 + 3 - x = -2x^2 + 5x + 3
\]

Right-hand side:
\[
5x(1 + 4x) = 5x + 20x^2
\]

Equate both sides:
\[
-2x^2 + 5x + 3 = 5x + 20x^2
\]

Simplify:
\[
-2x^2 + 3 = 20x^2
\]

Move all terms to one side:
\[
-22x^2 + 3 = 0
\]

Solve for \(x^2\):
\[
22x^2 = 3 \quad \Rightarrow \quad x^2 = \frac{3}{22}
\]

So, \(x = \pm \sqrt{\frac{3}{22}}\).

d) \( (3x + 7)^2 = 2 \)

Step 1: Take the square root of both sides.
\[
3x + 7 = \pm \sqrt{2}
\]

Step 2: Solve for \(x\):
\[
3x = -7 \pm \sqrt{2} \quad \Rightarrow \quad x = \frac{-7 \pm \sqrt{2}}{3}
\]

So, the solutions are \(x = \frac{-7 + \sqrt{2}}{3}\) and \(x = \frac{-7 - \sqrt{2}}{3}\).

#### 3) A, B et C sont trois points tels que:
\[
AB = \sqrt{543}, \quad AC = \sqrt{700}, \quad BC = \sqrt{63}
\]
Les points \(A\), \(B\), et \(C\) sont-ils alignés ?

To check if the points are aligned, use the triangle inequality. If the sum of two sides equals the third side, the points are aligned.

Check if:
\[
AB + BC = AC
\]
Substitute the values:
\[
\sqrt{543} + \sqrt{63} = \sqrt{700}
\]
Calculate:
\[
\sqrt{543} + \sqrt{63} = 23.30 + 7.94 = 31.24
\]
\[
\sqrt{700} = 26.46
\]
Since \(31.24 \neq 26.46\), the points are **not aligned**.

---

Would you like more details or have any other questions? Here are five related questions:

1. How do you solve equations with square roots?
2. How do you expand and simplify binomials?
3. What is the method to check if points are aligned using distances?
4. How does the triangle inequality theorem work in coordinate geometry?
5. How do conjugate pairs help in simplifying expressions with square roots?

**Tip**: Always check your expansion steps carefully to avoid missing any terms, especially when dealing with binomial squares.

Solve algebraic expressions and equations, including the expansion of binomials, solving quadratic and linear equations, and checking if points are aligned based on distances.

This exercise covers the expansion and simplification of binomials, solving quadratic and linear equations, and determining if three points are aligned based on their distances. Detailed solutions include algebraic techniques and geometry concepts, suitable for high school students in Grades 10-12.

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