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Sagot :
Sure, let's solve the inequality [tex]\(\frac{3}{2-x} \leq \frac{1}{x+4}\)[/tex] step-by-step.
### 1. Combine the Fractions
First, let's get both terms on one side and combine the fractions.
[tex]\[ \frac{3}{2-x} - \frac{1}{x+4} \leq 0 \][/tex]
To combine these fractions, we need a common denominator, which is [tex]\((2-x)(x+4)\)[/tex].
[tex]\[ \frac{3(x+4)}{(2-x)(x+4)} - \frac{1(2-x)}{(2-x)(x+4)} \leq 0 \][/tex]
Combine the fractions into one:
[tex]\[ \frac{3(x+4) - (2-x)}{(2-x)(x+4)} \leq 0 \][/tex]
### 2. Simplify the Numerator
Now, simplify the numerator:
[tex]\[ 3(x+4) - (2-x) = 3x + 12 - 2 + x = 4x + 10 \][/tex]
So, the inequality now is:
[tex]\[ \frac{4x + 10}{(2-x)(x+4)} \leq 0 \][/tex]
### 3. Find Critical Points
Next, identify the points where the numerator and denominator are zero because these points are where the expression might change its sign.
- The numerator [tex]\(4x + 10 = 0\)[/tex] at [tex]\(x = -\frac{5}{2}\)[/tex].
- The denominator [tex]\((2-x)(x+4) = 0\)[/tex] at [tex]\(x = 2\)[/tex] and [tex]\(x = -4\)[/tex].
Additionally, note that at [tex]\(x = 2\)[/tex] and [tex]\(x = -4\)[/tex], the expression is undefined since the denominator becomes zero.
### 4. Sign Analysis
Now, we will analyze the sign of the expression over the intervals defined by these critical points: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, -\frac{5}{2}) \)[/tex], [tex]\( (-\frac{5}{2}, 2) \)[/tex], and [tex]\( (2, \infty) \)[/tex].
1. Interval [tex]\((-∞, -4)\)[/tex]:
- Pick [tex]\(x = -5\)[/tex]:
[tex]\[ \frac{4(-5) + 10}{(2-(-5))(x+4)} = \frac{-20 + 10}{7(-1)} = \frac{-10}{-7} > 0 \][/tex]
2. Interval [tex]\((-4, -\frac{5}{2})\)[/tex]:
- Pick [tex]\(x = -3\)[/tex]:
[tex]\[ \frac{4(-3) + 10}{(2-(-3))(x+4)} = \frac{-12 + 10}{5(1)} = \frac{-2}{5} < 0 \][/tex]
3. Interval [tex]\((- \frac{5}{2}, 2)\)[/tex]:
- Pick [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{4(0) + 10}{(2-0)(0+4)} = \frac{10}{8} > 0 \][/tex]
4. Interval [tex]\((2, ∞)\)[/tex]:
- Pick [tex]\(x = 3\)[/tex]:
[tex]\[ \frac{4(3) + 10}{(2-3)(3+4)} = \frac{12 + 10}{-1(7)} = \frac{22}{-7} < 0 \][/tex]
### 5. Solution
We want the expression to be less than or equal to zero:
So, look at intervals where the expression is less than or equal to zero:
- It is negative (true for ≤ 0) in the intervals [tex]\((-4, -\frac{5}{2}) \cup (2, ∞)\)[/tex].
Lastly, check the points where the expression equals zero:
- The numerator is zero at [tex]\(x = -\frac{5}{2}\)[/tex]. We include this point.
- The points [tex]\(x = -4\)[/tex] and [tex]\(x = 2\)[/tex] are where the denominator is zero, making the expression undefined. These points are not included.
Hence, the solution to the inequality is:
[tex]\[ (-4, -\frac{5}{2}] \cup (2, \infty) \][/tex]
### 1. Combine the Fractions
First, let's get both terms on one side and combine the fractions.
[tex]\[ \frac{3}{2-x} - \frac{1}{x+4} \leq 0 \][/tex]
To combine these fractions, we need a common denominator, which is [tex]\((2-x)(x+4)\)[/tex].
[tex]\[ \frac{3(x+4)}{(2-x)(x+4)} - \frac{1(2-x)}{(2-x)(x+4)} \leq 0 \][/tex]
Combine the fractions into one:
[tex]\[ \frac{3(x+4) - (2-x)}{(2-x)(x+4)} \leq 0 \][/tex]
### 2. Simplify the Numerator
Now, simplify the numerator:
[tex]\[ 3(x+4) - (2-x) = 3x + 12 - 2 + x = 4x + 10 \][/tex]
So, the inequality now is:
[tex]\[ \frac{4x + 10}{(2-x)(x+4)} \leq 0 \][/tex]
### 3. Find Critical Points
Next, identify the points where the numerator and denominator are zero because these points are where the expression might change its sign.
- The numerator [tex]\(4x + 10 = 0\)[/tex] at [tex]\(x = -\frac{5}{2}\)[/tex].
- The denominator [tex]\((2-x)(x+4) = 0\)[/tex] at [tex]\(x = 2\)[/tex] and [tex]\(x = -4\)[/tex].
Additionally, note that at [tex]\(x = 2\)[/tex] and [tex]\(x = -4\)[/tex], the expression is undefined since the denominator becomes zero.
### 4. Sign Analysis
Now, we will analyze the sign of the expression over the intervals defined by these critical points: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, -\frac{5}{2}) \)[/tex], [tex]\( (-\frac{5}{2}, 2) \)[/tex], and [tex]\( (2, \infty) \)[/tex].
1. Interval [tex]\((-∞, -4)\)[/tex]:
- Pick [tex]\(x = -5\)[/tex]:
[tex]\[ \frac{4(-5) + 10}{(2-(-5))(x+4)} = \frac{-20 + 10}{7(-1)} = \frac{-10}{-7} > 0 \][/tex]
2. Interval [tex]\((-4, -\frac{5}{2})\)[/tex]:
- Pick [tex]\(x = -3\)[/tex]:
[tex]\[ \frac{4(-3) + 10}{(2-(-3))(x+4)} = \frac{-12 + 10}{5(1)} = \frac{-2}{5} < 0 \][/tex]
3. Interval [tex]\((- \frac{5}{2}, 2)\)[/tex]:
- Pick [tex]\(x = 0\)[/tex]:
[tex]\[ \frac{4(0) + 10}{(2-0)(0+4)} = \frac{10}{8} > 0 \][/tex]
4. Interval [tex]\((2, ∞)\)[/tex]:
- Pick [tex]\(x = 3\)[/tex]:
[tex]\[ \frac{4(3) + 10}{(2-3)(3+4)} = \frac{12 + 10}{-1(7)} = \frac{22}{-7} < 0 \][/tex]
### 5. Solution
We want the expression to be less than or equal to zero:
So, look at intervals where the expression is less than or equal to zero:
- It is negative (true for ≤ 0) in the intervals [tex]\((-4, -\frac{5}{2}) \cup (2, ∞)\)[/tex].
Lastly, check the points where the expression equals zero:
- The numerator is zero at [tex]\(x = -\frac{5}{2}\)[/tex]. We include this point.
- The points [tex]\(x = -4\)[/tex] and [tex]\(x = 2\)[/tex] are where the denominator is zero, making the expression undefined. These points are not included.
Hence, the solution to the inequality is:
[tex]\[ (-4, -\frac{5}{2}] \cup (2, \infty) \][/tex]
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