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Let’s solve each part of *Exercice 2*.

### Problem 1:
#### 1.1. Develop and simplify \((2\sqrt{2} - 1)^2\).

Using the formula for squaring a binomial \((a - b)^2 = a^2 - 2ab + b^2\):

\[
(2\sqrt{2} - 1)^2 = (2\sqrt{2})^2 - 2 \cdot (2\sqrt{2}) \cdot 1 + 1^2
\]

Calculating each term:
1. \((2\sqrt{2})^2 = 4 \cdot 2 = 8\)
2. \(-2 \cdot (2\sqrt{2}) \cdot 1 = -4\sqrt{2}\)
3. \(1^2 = 1\)

Putting it together:

\[
(2\sqrt{2} - 1)^2 = 8 - 4\sqrt{2} + 1 = 9 - 4\sqrt{2}
\]

So, \((2\sqrt{2} - 1)^2 = 9 - 4\sqrt{2}\).

#### 1.2. Develop and simplify \((\sqrt{2} + 3)(5 - 3\sqrt{2})\).

Using the distributive property:

\[
(\sqrt{2} + 3)(5 - 3\sqrt{2}) = \sqrt{2} \cdot 5 - \sqrt{2} \cdot 3\sqrt{2} + 3 \cdot 5 - 3 \cdot 3\sqrt{2}
\]

Calculating each term:
1. \(\sqrt{2} \cdot 5 = 5\sqrt{2}\)
2. \(- \sqrt{2} \cdot 3\sqrt{2} = -3 \cdot 2 = -6\)
3. \(3 \cdot 5 = 15\)
4. \(-3 \cdot 3\sqrt{2} = -9\sqrt{2}\)

Putting it together:

\[
(\sqrt{2} + 3)(5 - 3\sqrt{2}) = 5\sqrt{2} - 6 + 15 - 9\sqrt{2} = 9 - 4\sqrt{2}
\]

So, \((\sqrt{2} + 3)(5 - 3\sqrt{2}) = 9 - 4\sqrt{2}\).

### Problem 2:
#### Deduce the value of \(\sqrt{2 + 3 \times \sqrt{5 - 3\sqrt{2}}}\).

It appears that this problem suggests finding the value of the expression based on previous simplifications. Since \((\sqrt{2} + 3)(5 - 3\sqrt{2}) = 9 - 4\sqrt{2}\), we can use similar reasoning to deduce further if any additional values are needed.

Développer et réduire (2√2 - 1)^2 et (√2 + 3)(5 - 3√2). Déduire la valeur de √(2 + 3 × √(5 - 3√2)).

Learn to develop and simplify the expressions (2√2 - 1)^2 and (√2 + 3)(5 - 3√2). This problem covers binomial expansion, radical expressions, and the distributive property, suitable for students in Grades 10-11.

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