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Sagot :
Let's solve the given inequality step by step.
Step 1: Express on a common denominator
We start with the given inequality:
[tex]\[ \frac{x+3}{x-4} > \frac{x-1}{x+6} \][/tex]
To combine the fractions, we find a common denominator, which is [tex]\((x-4)(x+6)\)[/tex].
Rewriting each fraction:
[tex]\[ \frac{(x+3)(x+6)}{(x-4)(x+6)} > \frac{(x-1)(x-4)}{(x-4)(x+6)} \][/tex]
Since the denominators are the same, we can compare the numerators:
[tex]\[ \frac{(x+3)(x+6)}{(x-4)(x+6)} - \frac{(x-1)(x-4)}{(x-4)(x+6)} > 0 \][/tex]
[tex]\[ \frac{(x+3)(x+6) - (x-1)(x-4)}{(x-4)(x+6)} > 0 \][/tex]
Step 2: Simplify the numerator
We need to simplify the expression in the numerator:
[tex]\[ (x+3)(x+6) - (x-1)(x-4) \][/tex]
Expand both terms:
[tex]\[ (x^2 + 9x + 18) - (x^2 - 5x + 4) \][/tex]
Combine like terms:
[tex]\[ x^2 + 9x + 18 - x^2 + 5x - 4 \][/tex]
[tex]\[ 14x + 14 \][/tex]
Factor out the common factor:
[tex]\[ 14(x + 1) \][/tex]
So our inequality is now:
[tex]\[ \frac{14(x+1)}{(x-4)(x+6)} > 0 \][/tex]
Step 3: Determine critical points
Next, find the critical points by setting the numerator and denominator to zero:
[tex]\[ 14(x + 1) = 0 \quad \Rightarrow \quad x = -1 \][/tex]
[tex]\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \][/tex]
[tex]\[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \][/tex]
The critical points are [tex]\( x = -1 \)[/tex], [tex]\( x = 4 \)[/tex], and [tex]\( x = -6 \)[/tex].
Step 4: Test intervals around the critical points
We divide the number line into intervals using these critical points and test each interval to determine where the inequality holds:
Intervals: [tex]\( (-\infty, -6) \)[/tex], [tex]\( (-6, -1) \)[/tex], [tex]\( (-1, 4) \)[/tex], [tex]\( (4, \infty) \)[/tex]
Choose test points in each interval:
1. [tex]\( x = -7 \)[/tex]:
[tex]\[ \frac{14(-7+1)}{(-7-4)(-7+6)} = \frac{14(-6)}{(-11)(-1)} = \frac{-84}{11} < 0 \][/tex]
2. [tex]\( x = -2 \)[/tex]:
[tex]\[ \frac{14(-2+1)}{(-2-4)(-2+6)} = \frac{14(-1)}{(-6)(4)} = \frac{-14}{-24} > 0 \][/tex]
3. [tex]\( x = 0 \)[/tex]:
[tex]\[ \frac{14(0+1)}{(0-4)(0+6)} = \frac{14(1)}{(-4)(6)} = \frac{14}{-24} < 0 \][/tex]
4. [tex]\( x = 5 \)[/tex]:
[tex]\[ \frac{14(5+1)}{(5-4)(5+6)} = \frac{14(6)}{(1)(11)} = \frac{84}{11} > 0 \][/tex]
Step 5: Assemble solution intervals
The inequality [tex]\(\frac{14(x+1)}{(x-4)(x+6)} > 0\)[/tex] is satisfied in the intervals [tex]\( (-6, -1) \)[/tex] and [tex]\( (4, \infty) \)[/tex].
Step 6: State in interval notation
Since the points [tex]\( x = -6 \)[/tex] and [tex]\( x = 4 \)[/tex] make the denominator zero, and [tex]\( x = -1 \)[/tex] makes the numerator zero, we exclude these points from our solution:
Thus, the solution in interval notation is:
[tex]\[ (-6, -1) \cup (4, \infty) \][/tex]
Step 1: Express on a common denominator
We start with the given inequality:
[tex]\[ \frac{x+3}{x-4} > \frac{x-1}{x+6} \][/tex]
To combine the fractions, we find a common denominator, which is [tex]\((x-4)(x+6)\)[/tex].
Rewriting each fraction:
[tex]\[ \frac{(x+3)(x+6)}{(x-4)(x+6)} > \frac{(x-1)(x-4)}{(x-4)(x+6)} \][/tex]
Since the denominators are the same, we can compare the numerators:
[tex]\[ \frac{(x+3)(x+6)}{(x-4)(x+6)} - \frac{(x-1)(x-4)}{(x-4)(x+6)} > 0 \][/tex]
[tex]\[ \frac{(x+3)(x+6) - (x-1)(x-4)}{(x-4)(x+6)} > 0 \][/tex]
Step 2: Simplify the numerator
We need to simplify the expression in the numerator:
[tex]\[ (x+3)(x+6) - (x-1)(x-4) \][/tex]
Expand both terms:
[tex]\[ (x^2 + 9x + 18) - (x^2 - 5x + 4) \][/tex]
Combine like terms:
[tex]\[ x^2 + 9x + 18 - x^2 + 5x - 4 \][/tex]
[tex]\[ 14x + 14 \][/tex]
Factor out the common factor:
[tex]\[ 14(x + 1) \][/tex]
So our inequality is now:
[tex]\[ \frac{14(x+1)}{(x-4)(x+6)} > 0 \][/tex]
Step 3: Determine critical points
Next, find the critical points by setting the numerator and denominator to zero:
[tex]\[ 14(x + 1) = 0 \quad \Rightarrow \quad x = -1 \][/tex]
[tex]\[ x - 4 = 0 \quad \Rightarrow \quad x = 4 \][/tex]
[tex]\[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \][/tex]
The critical points are [tex]\( x = -1 \)[/tex], [tex]\( x = 4 \)[/tex], and [tex]\( x = -6 \)[/tex].
Step 4: Test intervals around the critical points
We divide the number line into intervals using these critical points and test each interval to determine where the inequality holds:
Intervals: [tex]\( (-\infty, -6) \)[/tex], [tex]\( (-6, -1) \)[/tex], [tex]\( (-1, 4) \)[/tex], [tex]\( (4, \infty) \)[/tex]
Choose test points in each interval:
1. [tex]\( x = -7 \)[/tex]:
[tex]\[ \frac{14(-7+1)}{(-7-4)(-7+6)} = \frac{14(-6)}{(-11)(-1)} = \frac{-84}{11} < 0 \][/tex]
2. [tex]\( x = -2 \)[/tex]:
[tex]\[ \frac{14(-2+1)}{(-2-4)(-2+6)} = \frac{14(-1)}{(-6)(4)} = \frac{-14}{-24} > 0 \][/tex]
3. [tex]\( x = 0 \)[/tex]:
[tex]\[ \frac{14(0+1)}{(0-4)(0+6)} = \frac{14(1)}{(-4)(6)} = \frac{14}{-24} < 0 \][/tex]
4. [tex]\( x = 5 \)[/tex]:
[tex]\[ \frac{14(5+1)}{(5-4)(5+6)} = \frac{14(6)}{(1)(11)} = \frac{84}{11} > 0 \][/tex]
Step 5: Assemble solution intervals
The inequality [tex]\(\frac{14(x+1)}{(x-4)(x+6)} > 0\)[/tex] is satisfied in the intervals [tex]\( (-6, -1) \)[/tex] and [tex]\( (4, \infty) \)[/tex].
Step 6: State in interval notation
Since the points [tex]\( x = -6 \)[/tex] and [tex]\( x = 4 \)[/tex] make the denominator zero, and [tex]\( x = -1 \)[/tex] makes the numerator zero, we exclude these points from our solution:
Thus, the solution in interval notation is:
[tex]\[ (-6, -1) \cup (4, \infty) \][/tex]
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