https://file.aiquickdraw.com/mathbot/file/1727010093256f35xrn2g.jpg

Let's analyze and solve each of the propositions presented in the image. Here's a step-by-step breakdown:

### **Exercise 1:**

We need to determine which of the propositions \((P_1) \) to \((P_7) \) are true or false and justify them.

#### **1. Proposition \( P_1 \):**

\[
(\forall x \in \mathbb{R}) : -6x^2 + 5x - 1 \leq 0
\]

Let's solve the quadratic inequality:
The expression is \( -6x^2 + 5x - 1 \). 
We find the roots of the equation \( -6x^2 + 5x - 1 = 0 \) using the discriminant method:
\[
b^2 - 4ac = 5^2 - 4(-6)(-1) = 25 - 24 = 1
\]
The roots are \( x = 1 \) and \( x = \frac{1}{6} \).

For \( x \in \left[\frac{1}{6}, 1\right] \), \( -6x^2 + 5x - 1 \) will be less than or equal to 0. However, for other \( x \), the expression will be positive. Thus, \( P_1 \) is false.

#### **2. Proposition \( P_2 \):**

\[
(\exists! x \in \mathbb{R}) : x^3 - 2\sqrt{2} = 0
\]
Solving, \( x^3 = 2\sqrt{2} \), so \( x = \sqrt[3]{2\sqrt{2}} \). There exists exactly one real solution to this equation. Hence, \( P_2 \) is true.

#### **3. Proposition \( P_3 \):**

\[
(\forall m \in \mathbb{R}^*)(\exists x \in \mathbb{R}) : x^2 - mx + m^2 = 0
\]

The discriminant of this quadratic equation is \( m^2 - 4m^2 = -3m^2 \). For \( m \neq 0 \), this discriminant is always negative, implying no real solutions. Therefore, \( P_3 \) is false.

#### **4. Proposition \( P_4 \):**

\[
(\forall m \in \mathbb{R})(\exists x \in \mathbb{R}) : x^2 - 3mx + 2m^2 + 1 = 0
\]

The discriminant here is \( (3m)^2 - 4(1)(2m^2 + 1) = 9m^2 - 8m^2 - 4 = m^2 - 4 \). 
For all \( m \neq \pm 2 \), there are real solutions. When \( m = 2 \) or \( m = -2 \), it has a double root. Hence, \( P_4 \) is true.

#### **5. Proposition \( P_5 \):**

\[
(\exists m \in \mathbb{R})(\exists x \in \mathbb{R}) : x^2 - 3mx + 2m^2 + 1 = 0
\]
As verified previously, for most values of \( m \), this has solutions, so \( P_5 \) is true.

#### **6. Proposition \( P_6 \):**

(a) Check \( m = 6 \), then \( x^2 + x + 5 = 6 \), which gives \( x^2 + x - 1 = 0 \). The discriminant is \( 1 + 4 = 5 > 0 \), so \( P_6 \) is true.

For \( m = -1 \), \( x^2 + x + 5 = -1 \) leads to a contradiction, so \( P_6 \) is false.

#### **7. Proposition \( P_7 \):**

Checking, you will find the propositions work for any \( m > 0 \).

### Do you want a more detailed explanation of any part? Here are some related questions:

1. What is the definition of discriminants and how do they determine the nature of solutions for quadratic equations?
2. Can you explain why \( P_1 \) is false by graphing the function \( -6x^2 + 5x - 1 \)?
3. Why do discriminants differ for equations in \( P_4 \) and \( P_5 \) cases?
4. What other real solutions exist for \( x^3 = 2\sqrt{2} \)?
5. How can uniqueness be verified in mathematical proofs like \( P_2 \)?

**Tip:** Always check your roots using a discriminant!

Distinguish the true propositions from the false ones and justify your answers for propositions P1 to P7.

This problem involves distinguishing true and false mathematical propositions using quadratic equations, discriminants, and inequalities. Learn how to justify solutions using real number properties and unique existence theorems.

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