Answer
### Exercice 1: Résolution des Équations et Polynômes
1. **Résolution des Équations et Inéquations**
**a. Équation :**
\[
\frac{2x - 1}{2} + \frac{3x - 2}{4} = 20 - \frac{5x}{7}
\]
- **Solution :**
\[
x = \frac{196}{23}
\]
**b. Équation :**
\[
|2x - 6| + |x + 6| = 5
\]
- **Solution :**
Aucune solution dans \(\mathbb{R}\).
**c. Équation :**
\[
2x^2 - 7x + 6 = 0
\]
- **Solutions :**
\[
x_1 = \frac{3}{2}, \quad x_2 = 2
\]
2. **Polynôme \( P(x) = 2x^3 - 5x^2 - x + 6 \)**
**a. Calcul de \( P(1) \) :**
\[
P(1) = 2(1)^3 - 5(1)^2 - 1 + 6 = 2 - 5 - 1 + 6 = 2
\]
**b. Calcul de \( P(-1) \) :**
\[
P(-1) = 2(-1)^3 - 5(-1)^2 - (-1) + 6 = -2 - 5 + 1 + 6 = 0
\]
**c. Résolution de \( P(x) = 0 \) :**
- **Solutions :**
\[
x_1 = -1, \quad x_2 = \frac{3}{2}, \quad x_3 = 2
\]
**d. Factorisation de \( P(x) \) :**
En utilisant la division euclidienne, \( P(x) \) peut être écrit comme :
\[
P(x) = (x + 1)(2x - 3)(x - 2)
\]
**e. Résolution de \( P(x) > 0 \) :**
- **Intervalle où \( P(x) > 0 \) :**
\[
x < -1 \quad \text{ou} \quad x > 2
\]
### Conclusion
Nous avons résolu les équations et inéquations proposées, calculé les valeurs du polynôme en \( x = 1 \) et \( x = -1 \), et factorisé le polynôme \( P(x) \).
Solution
Evaluate the expression by following steps:
- step0: Evaluate:
\(2x^{3}-5x^{2}-x+6\)
- step1: Substitute:
\(2\times 1^{3}-5\times 1^{2}-1+6\)
- step2: Evaluate the power:
\(2\times 1-5\times 1^{2}-1+6\)
- step3: Evaluate the power:
\(2\times 1-5\times 1-1+6\)
- step4: Multiply:
\(2-5\times 1-1+6\)
- step5: Multiply:
\(2-5-1+6\)
- step6: Calculate:
\(2\)
Solve the equation \( 2 x^{2}-7 x+6=0 \).
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
\(2x^{2}-7x+6=0\)
- step1: Factor the expression:
\(\left(x-2\right)\left(2x-3\right)=0\)
- step2: Separate into possible cases:
\(\begin{align}&x-2=0\\&2x-3=0\end{align}\)
- step3: Solve the equation:
\(\begin{align}&x=2\\&x=\frac{3}{2}\end{align}\)
- step4: Rewrite:
\(x_{1}=\frac{3}{2},x_{2}=2\)
Substitute \( x=-1 \) into the expression \( 2 x^{3}-5 x^{2}-x+6 \).
Evaluate the expression by following steps:
- step0: Evaluate:
\(2x^{3}-5x^{2}-x+6\)
- step1: Substitute:
\(2\left(-1\right)^{3}-5\left(-1\right)^{2}-\left(-1\right)+6\)
- step2: Evaluate the power:
\(2\left(-1\right)^{3}-5\times 1-\left(-1\right)+6\)
- step3: Multiply the terms:
\(-2-5\times 1-\left(-1\right)+6\)
- step4: Multiply:
\(-2-5-\left(-1\right)+6\)
- step5: Subtract the numbers:
\(-7-\left(-1\right)+6\)
- step6: Subtract the terms:
\(-6+6\)
- step7: Add the numbers:
\(0\)
Solve the equation \( \frac{2 x-1}{2}+\frac{3 x-2}{4}=20-\frac{5 x}{7} \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\frac{2x-1}{2}+\frac{3x-2}{4}=20-\frac{5x}{7}\)
- step1: Multiply both sides of the equation by LCD:
\(\left(\frac{2x-1}{2}+\frac{3x-2}{4}\right)\times 28=\left(20-\frac{5x}{7}\right)\times 28\)
- step2: Simplify the equation:
\(49x-28=560-20x\)
- step3: Move the expression to the left side:
\(49x+20x=560+28\)
- step4: Add and subtract:
\(69x=560+28\)
- step5: Add and subtract:
\(69x=588\)
- step6: Divide both sides:
\(\frac{69x}{69}=\frac{588}{69}\)
- step7: Divide the numbers:
\(x=\frac{196}{23}\)
Solve the equation \( |2 x-6|+|x+6|=5 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(\left|2x-6\right|+\left|x+6\right|=5\)
- step1: Move the expression to the left side:
\(\left|2x-6\right|+\left|x+6\right|-5=0\)
- step2: Separate into possible cases:
\(\begin{align}&2x-6+x+6-5=0,2x-6\geq 0,x+6\geq 0\\&2x-6-\left(x+6\right)-5=0,2x-6\geq 0,x+6<0\\&-\left(2x-6\right)+x+6-5=0,2x-6<0,x+6\geq 0\\&-\left(2x-6\right)-\left(x+6\right)-5=0,2x-6<0,x+6<0\end{align}\)
- step3: Solve the equation:
\(\begin{align}&x=\frac{5}{3},x\geq 3,x\geq -6\\&x=17,x\geq 3,x<-6\\&x=7,x<3,x\geq -6\\&x=-\frac{5}{3},x<3,x<-6\end{align}\)
- step4: Find the intersection:
\(\begin{align}&x \in \varnothing \\&x \in \varnothing \\&x \in \varnothing \\&x \in \varnothing \end{align}\)
- step5: Find the union:
\(x \in \varnothing \)
Solve the equation \( 2 x^{3}-5 x^{2}-x+6=0 \).
Solve the equation by following steps:
- step0: Solve for \(x\):
\(2x^{3}-5x^{2}-x+6=0\)
- step1: Factor the expression:
\(\left(x-2\right)\left(x+1\right)\left(2x-3\right)=0\)
- step2: Separate into possible cases:
\(\begin{align}&x-2=0\\&x+1=0\\&2x-3=0\end{align}\)
- step3: Solve the equation:
\(\begin{align}&x=2\\&x=-1\\&x=\frac{3}{2}\end{align}\)
- step4: Rewrite:
\(x_{1}=-1,x_{2}=\frac{3}{2},x_{3}=2\)
Expand the expression \( 2 x^{3}-5 x^{2}-x+6 \)
Simplify the expression by following steps:
- step0: :
\(2x^{3}-5x^{2}-x+6\)
Solve the equation \( P(x)>0 \).
Solve the inequality by following steps:
- step0: Solve for \(P,x\):
\(Px>0\)
- step1: Separate into possible cases:
\(\begin{align}&\left\{ \begin{array}{l}P>0\\x>0\end{array}\right.\\&\left\{ \begin{array}{l}P<0\\x<0\end{array}\right.\end{align}\)
- step2: Find the union:
\(\begin{align}&\left\{ \begin{array}{l}P<0\\x<0\end{array}\right.\\&\left\{ \begin{array}{l}P>0\\x>0\end{array}\right.\end{align}\)
### Exercice 1: Résolution des équations et polynômes
#### 1. Résoudre les équations et inéquations
**a. Équation :**
\[
\frac{2 x-1}{2}+\frac{3 x-2}{4}=20-\frac{5 x}{7}
\]
**Solution :**
En résolvant cette équation, nous trouvons :
\[
x = \frac{196}{23}
\]
---
**b. Équation :**
\[
|2 x-6|+|x+6|=5
\]
**Solution :**
Cette équation n'a pas de solution dans \(\mathbb{R}\) :
\[
x \in \varnothing
\]
---
**c. Équation :**
\[
2 x^{2}-7 x+6=0
\]
**Solution :**
Les solutions de cette équation sont :
\[
x_{1} = \frac{3}{2}, \quad x_{2} = 2
\]
---
#### 2. Polynôme \( P(x) = 2 x^{3}-5 x^{2}-x+6 \)
**a. Calculer \( P(1) \) :**
\[
P(1) = 2(1)^{3} - 5(1)^{2} - (1) + 6 = 2 - 5 - 1 + 6 = 2
\]
**b. Calculer \( P(-1) \) :**
\[
P(-1) = 2(-1)^{3} - 5(-1)^{2} - (-1) + 6 = -2 - 5 + 1 + 6 = 0
\]
**c. Résoudre \( P(x) = 0 \) :**
Les solutions de l'équation \( P(x) = 0 \) sont :
\[
x_{1} = -1, \quad x_{2} = \frac{3}{2}, \quad x_{3} = 2
\]
**d. Factoriser \( P(x) \) :**
En utilisant la division euclidienne, nous pouvons écrire \( P(x) \) sous forme de produit de binômes.
**e. Résoudre \( P(x) > 0 \) :**
Les intervalles où \( P(x) > 0 \) sont :
\[
\left\{ P < 0; x < 0 \right\}; \quad \left\{ P > 0; x > 0 \right\}
\]
### Conclusion
Nous avons résolu les équations et inéquations demandées, calculé les valeurs du polynôme en \( x = 1 \) et \( x = -1 \), et factorisé le polynôme \( P(x) \).
Answered by UpStudy AI and reviewed by a Professional Tutor
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Explain
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Simplify this solution