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Sagot :
To solve the inequality [tex]\( |x - \sqrt{2}| > |x + 3\sqrt{2}| \)[/tex], let's examine the different cases based on the properties of absolute values. We know that the expression inside an absolute value changes depending on whether the argument is positive or negative.
### Step 1: Consider Different Cases
We'll analyze the inequality [tex]\( |x - \sqrt{2}| > |x + 3\sqrt{2}| \)[/tex] by breaking it down into several cases based on the values of [tex]\( x \)[/tex] relative to [tex]\( \sqrt{2} \)[/tex] and [tex]\( -3\sqrt{2} \)[/tex]:
1. [tex]\( x \geq \sqrt{2} \)[/tex]
2. [tex]\( -3\sqrt{2} \leq x < \sqrt{2} \)[/tex]
3. [tex]\( x < -3\sqrt{2} \)[/tex]
### Case 1: [tex]\( x \geq \sqrt{2} \)[/tex]
In this case:
- [tex]\( |x - \sqrt{2}| = x - \sqrt{2} \)[/tex]
- [tex]\( |x + 3\sqrt{2}| = x + 3\sqrt{2} \)[/tex]
So, the inequality becomes:
[tex]\[ x - \sqrt{2} > x + 3\sqrt{2} \][/tex]
Simplifying, we get:
[tex]\[ -\sqrt{2} > 3\sqrt{2} \][/tex]
[tex]\[ -4\sqrt{2} > 0 \][/tex]
This inequality is not true. Thus, there is no solution in this case.
### Case 2: [tex]\( -3\sqrt{2} \leq x < \sqrt{2} \)[/tex]
In this case:
- [tex]\( |x - \sqrt{2}| = \sqrt{2} - x \)[/tex] (since [tex]\( x < \sqrt{2} \)[/tex])
- [tex]\( |x + 3\sqrt{2}| = x + 3\sqrt{2} \)[/tex] (since [tex]\( x\)[/tex] is non-negative)
So, the inequality becomes:
[tex]\[ \sqrt{2} - x > x + 3\sqrt{2} \][/tex]
Simplifying, we get:
[tex]\[ \sqrt{2} - 3\sqrt{2} > x + x \][/tex]
[tex]\[ -2\sqrt{2} > 2x \][/tex]
[tex]\[ -\sqrt{2} > x \][/tex]
And since we know from this case criterion that [tex]\( x \geq -3\sqrt{2} \)[/tex], we specifically consider the valid range for [tex]\( x \)[/tex]:
[tex]\[ -\sqrt{2} \leq x < \sqrt{2}\][/tex]
Hence this inequality is satisfied for [tex]\( x < -\sqrt{2} \)[/tex].
### Case 3: [tex]\( x < -3\sqrt{2} \)[/tex]
In this case:
- [tex]\( |x - \sqrt{2}| = \sqrt{2} - x \)[/tex]
- [tex]\( |x + 3\sqrt{2}| = -(x + 3\sqrt{2}) = -x - 3\sqrt{2} \)[/tex] (since [tex]\( x + 3\sqrt{2} \)[/tex] is negative when [tex]\( x < -3\sqrt{2} \)[/tex])
So, the inequality becomes:
[tex]\[ \sqrt{2} - x > -x - 3\sqrt{2} \][/tex]
Simplifying, we get:
[tex]\[ \sqrt{2} - x > -x - 3\sqrt{2} \][/tex]
[tex]\[ \sqrt{2} > -3\sqrt{2} \][/tex]
Since this inequality is always true, any values in this range are solutions to the inequality.
### Conclusion:
From the analysis, the solutions to the inequality are:
- There are no solutions in [tex]\( x \geq \sqrt{2} \)[/tex].
- The inequality holds for all [tex]\( x < -\sqrt{2} \)[/tex].
Thus, the correct choice is [tex]\(\boxed{x < -\sqrt{2}}\)[/tex]:
(A) [tex]\( x < -\sqrt{2} \)[/tex]
### Step 1: Consider Different Cases
We'll analyze the inequality [tex]\( |x - \sqrt{2}| > |x + 3\sqrt{2}| \)[/tex] by breaking it down into several cases based on the values of [tex]\( x \)[/tex] relative to [tex]\( \sqrt{2} \)[/tex] and [tex]\( -3\sqrt{2} \)[/tex]:
1. [tex]\( x \geq \sqrt{2} \)[/tex]
2. [tex]\( -3\sqrt{2} \leq x < \sqrt{2} \)[/tex]
3. [tex]\( x < -3\sqrt{2} \)[/tex]
### Case 1: [tex]\( x \geq \sqrt{2} \)[/tex]
In this case:
- [tex]\( |x - \sqrt{2}| = x - \sqrt{2} \)[/tex]
- [tex]\( |x + 3\sqrt{2}| = x + 3\sqrt{2} \)[/tex]
So, the inequality becomes:
[tex]\[ x - \sqrt{2} > x + 3\sqrt{2} \][/tex]
Simplifying, we get:
[tex]\[ -\sqrt{2} > 3\sqrt{2} \][/tex]
[tex]\[ -4\sqrt{2} > 0 \][/tex]
This inequality is not true. Thus, there is no solution in this case.
### Case 2: [tex]\( -3\sqrt{2} \leq x < \sqrt{2} \)[/tex]
In this case:
- [tex]\( |x - \sqrt{2}| = \sqrt{2} - x \)[/tex] (since [tex]\( x < \sqrt{2} \)[/tex])
- [tex]\( |x + 3\sqrt{2}| = x + 3\sqrt{2} \)[/tex] (since [tex]\( x\)[/tex] is non-negative)
So, the inequality becomes:
[tex]\[ \sqrt{2} - x > x + 3\sqrt{2} \][/tex]
Simplifying, we get:
[tex]\[ \sqrt{2} - 3\sqrt{2} > x + x \][/tex]
[tex]\[ -2\sqrt{2} > 2x \][/tex]
[tex]\[ -\sqrt{2} > x \][/tex]
And since we know from this case criterion that [tex]\( x \geq -3\sqrt{2} \)[/tex], we specifically consider the valid range for [tex]\( x \)[/tex]:
[tex]\[ -\sqrt{2} \leq x < \sqrt{2}\][/tex]
Hence this inequality is satisfied for [tex]\( x < -\sqrt{2} \)[/tex].
### Case 3: [tex]\( x < -3\sqrt{2} \)[/tex]
In this case:
- [tex]\( |x - \sqrt{2}| = \sqrt{2} - x \)[/tex]
- [tex]\( |x + 3\sqrt{2}| = -(x + 3\sqrt{2}) = -x - 3\sqrt{2} \)[/tex] (since [tex]\( x + 3\sqrt{2} \)[/tex] is negative when [tex]\( x < -3\sqrt{2} \)[/tex])
So, the inequality becomes:
[tex]\[ \sqrt{2} - x > -x - 3\sqrt{2} \][/tex]
Simplifying, we get:
[tex]\[ \sqrt{2} - x > -x - 3\sqrt{2} \][/tex]
[tex]\[ \sqrt{2} > -3\sqrt{2} \][/tex]
Since this inequality is always true, any values in this range are solutions to the inequality.
### Conclusion:
From the analysis, the solutions to the inequality are:
- There are no solutions in [tex]\( x \geq \sqrt{2} \)[/tex].
- The inequality holds for all [tex]\( x < -\sqrt{2} \)[/tex].
Thus, the correct choice is [tex]\(\boxed{x < -\sqrt{2}}\)[/tex]:
(A) [tex]\( x < -\sqrt{2} \)[/tex]
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