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Déterminer la vérité par raisonnement d'équivalence 

Let's examine the given statement and prove it by reasoning with equivalences.

### Problem Statement
We have two positive real numbers \(a\) and \(b\) such that:
\[
a + b + 2 = 2\sqrt{a} + 2\sqrt{b}
\]
We need to show that this equation implies:
\[
a = b = 1
\]

### Solution
Let's proceed by simplifying and analyzing the given equation step-by-step.

1. **Rewrite the Equation**: 
   \[
   a + b + 2 = 2\sqrt{a} + 2\sqrt{b}
   \]
   Divide both sides by 2 to simplify:
   \[
   \frac{a}{2} + \frac{b}{2} + 1 = \sqrt{a} + \sqrt{b}
   \]

2. **Introduce New Variables**:
   Let \(x = \sqrt{a}\) and \(y = \sqrt{b}\), where \(x, y > 0\). Then \(a = x^2\) and \(b = y^2\), and we can rewrite the equation as:
   \[
   x^2 + y^2 + 2 = 2x + 2y
   \]
   Simplifying further:
   \[
   x^2 - 2x + y^2 - 2y + 2 = 0
   \]

3. **Complete the Square**:
   - For the \(x\)-terms: \(x^2 - 2x = (x - 1)^2 - 1\)
   - For the \(y\)-terms: \(y^2 - 2y = (y - 1)^2 - 1\)

   Substitute these into the equation:
   \[
   (x - 1)^2 - 1 + (y - 1)^2 - 1 + 2 = 0
   \]
   Simplify by combining constants:
   \[
   (x - 1)^2 + (y - 1)^2 = 0
   \]

4. **Analyze the Result**:
   Since \((x - 1)^2\) and \((y - 1)^2\) are both squares, they are non-negative. The only way their sum can be zero is if both terms are zero:
   \[
   (x - 1)^2 = 0 \quad \text{and} \quad (y - 1)^2 = 0
   \]
   Therefore, \(x = 1\) and \(y = 1\).

5. **Conclusion**:
   Since \(x = \sqrt{a} = 1\) and \(y = \sqrt{b} = 1\), we have \(a = 1\) and \(b = 1\).

### Final Answer
Thus, we have shown that \(a = b = 1\) is the only solution that satisfies the equation:
\[
a + b + 2 = 2\sqrt{a} + 2\sqrt{b} \iff a = b = 1
\]

Would you like further details on any step? 

Here are some related questions that expand on this topic:
1. How does completing the square help in solving equations involving squares?
2. Why is it necessary to use variables like \(x\) and \(y\) for square roots in proofs?
3. What are alternative methods to solve similar equations without completing the square?
4. Can you apply similar reasoning to inequalities, such as \(a + b + 2 > 2\sqrt{a} + 2\sqrt{b}\)?
5. How can you verify solutions by substituting them back into the original equation?

**Tip**: When you encounter square root terms, substituting with variables for simplicity (like \(x = \sqrt{a}\)) can make it easier to solve the equation systematically.

Soit (a, b) ∈ R+ × R+. Montrer que : a + b + 2 = 2√a + 2√b ⇔ a = b = 1

Explore how to prove that a + b + 2 = 2√a + 2√b implies a = b = 1. This algebraic problem involves concepts like real numbers, square roots, and completing the square, suitable for Grades 10-12.

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