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Déterminer la vérité par raisonnement de logique à une méthode facile

To solve this problem, we need to show that the equation:

\[
x^2 + y^2 + z^2 - ax - by - cz + d = 0
\]

is satisfied when \( x = \frac{a}{2} \), \( y = \frac{b}{2} \), and \( z = \frac{c}{2} \). Here’s a step-by-step approach:

### Step 1: Substitute \( x = \frac{a}{2} \), \( y = \frac{b}{2} \), and \( z = \frac{c}{2} \) into the equation.

1. Substitute these values for \(x\), \(y\), and \(z\) into the left-hand side of the equation:
   \[
   \left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 + \left(\frac{c}{2}\right)^2 - a\left(\frac{a}{2}\right) - b\left(\frac{b}{2}\right) - c\left(\frac{c}{2}\right) + d
   \]

### Step 2: Simplify each term individually.

1. Calculate the squares:
   \[
   \left(\frac{a}{2}\right)^2 = \frac{a^2}{4}, \quad \left(\frac{b}{2}\right)^2 = \frac{b^2}{4}, \quad \left(\frac{c}{2}\right)^2 = \frac{c^2}{4}
   \]
   Therefore, the sum of squares is:
   \[
   \frac{a^2}{4} + \frac{b^2}{4} + \frac{c^2}{4}
   \]

2. Calculate the linear terms:
   \[
   - a\left(\frac{a}{2}\right) = -\frac{a^2}{2}, \quad - b\left(\frac{b}{2}\right) = -\frac{b^2}{2}, \quad - c\left(\frac{c}{2}\right) = -\frac{c^2}{2}
   \]
   Summing these, we get:
   \[
   -\frac{a^2}{2} - \frac{b^2}{2} - \frac{c^2}{2}
   \]

### Step 3: Combine terms.

Combine the squared and linear terms:
\[
\frac{a^2}{4} + \frac{b^2}{4} + \frac{c^2}{4} - \frac{a^2}{2} - \frac{b^2}{2} - \frac{c^2}{2}
\]

To simplify, rewrite \(-\frac{a^2}{2}\) as \(-\frac{2a^2}{4}\) and do the same for the other terms:
\[
\left(\frac{a^2}{4} - \frac{2a^2}{4}\right) + \left(\frac{b^2}{4} - \frac{2b^2}{4}\right) + \left(\frac{c^2}{4} - \frac{2c^2}{4}\right)
\]
This reduces to:
\[
-\frac{a^2}{4} - \frac{b^2}{4} - \frac{c^2}{4}
\]

### Step 4: Final expression

So, the left-hand side of the equation becomes:
\[
-\frac{a^2}{4} - \frac{b^2}{4} - \frac{c^2}{4} + d
\]
For this to equal zero, we must have:
\[
d = \frac{a^2}{4} + \frac{b^2}{4} + \frac{c^2}{4}
\]

### Conclusion

Thus, the equation \( x^2 + y^2 + z^2 - ax - by - cz + d = 0 \) is satisfied if \( d = \frac{a^2}{4} + \frac{b^2}{4} + \frac{c^2}{4} \).

Would you like further details on each step, or do you have any questions?

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Here are five related questions:
1. What is the geometric interpretation of this equation in three-dimensional space?
2. How would the solution change if \(d\) had a different value?
3. Can this result be generalized to more variables?
4. How does this method relate to completing the square?
5. What if \(a\), \(b\), or \(c\) were negative?

**Tip:** Completing the square is often a useful technique in problems involving quadratic forms and can simplify solving equations like this.

Montrer que : x^2 + y^2 + z^2 - ax - by - cz + d = 0 ⟺ x = a/2 et y = b/2 et z = c/2

Learn how to demonstrate that x^2 + y^2 + z^2 - ax - by - cz + d = 0 holds true when x = a/2, y = b/2, and z = c/2. This problem covers algebraic techniques and completing the square, providing a step-by-step solution suitable for students in Grades 10-12.

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