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Sagot :
Let's analyze each of the given equations to determine which ones have exactly one solution.
### Equation (A)
[tex]\[ 12x + 12 = 13x - 12 \][/tex]
First, we'll isolate [tex]\(x\)[/tex] by moving all terms involving [tex]\(x\)[/tex] to one side and the constant terms to the other side:
[tex]\[ 12x + 12 = 13x - 12 \][/tex]
Subtract [tex]\(12x\)[/tex] from both sides:
[tex]\[ 12 = x - 12 \][/tex]
Add 12 to both sides:
[tex]\[ 24 = x \][/tex]
Therefore, equation (A) has exactly one solution, [tex]\(x = 24\)[/tex].
### Equation (B)
[tex]\[ -13x + 12 = 13x + 13 \][/tex]
Again, we'll isolate [tex]\(x\)[/tex] by moving all terms involving [tex]\(x\)[/tex] to one side and the constant terms to the other side:
[tex]\[ -13x + 12 = 13x + 13 \][/tex]
Add [tex]\(13x\)[/tex] to both sides:
[tex]\[ 12 = 26x + 13 \][/tex]
Subtract 13 from both sides:
[tex]\[ -1 = 26x \][/tex]
Divide both sides by 26:
[tex]\[ x = -\frac{1}{26} \][/tex]
Therefore, equation (B) has exactly one solution, [tex]\(x = -\frac{1}{26}\)[/tex].
### Equation (C)
[tex]\[ -13x + 12 = 13x - 13 \][/tex]
Isolating [tex]\(x\)[/tex], move all terms involving [tex]\(x\)[/tex] to one side and the constants to the other side:
[tex]\[ -13x + 12 = 13x - 13 \][/tex]
Add [tex]\(13x\)[/tex] to both sides:
[tex]\[ 12 = 26x - 13 \][/tex]
Add 13 to both sides:
[tex]\[ 25 = 26x \][/tex]
Divide both sides by 26:
[tex]\[ x = \frac{25}{26} \][/tex]
Therefore, equation (C) has exactly one solution, [tex]\(x = \frac{25}{26}\)[/tex].
### Equation (D)
[tex]\[ 12x + 12 = 13x + 12 \][/tex]
Isolating [tex]\(x\)[/tex], move all terms involving [tex]\(x\)[/tex] to one side and the constants to the other side:
[tex]\[ 12x + 12 = 13x + 12 \][/tex]
Subtract [tex]\(12x\)[/tex] from both sides:
[tex]\[ 12 = x + 12 \][/tex]
Subtract 12 from both sides:
[tex]\[ 0 = x \][/tex]
Therefore, equation (D) has exactly one solution, [tex]\(x = 0\)[/tex].
### Conclusion
Upon reviewing the calculations, it is evident that each of the equations (A), (B), (C), and (D) has exactly one solution.
So, the correct answer is:
- (A)
- (B)
- (C)
- (D)
### Equation (A)
[tex]\[ 12x + 12 = 13x - 12 \][/tex]
First, we'll isolate [tex]\(x\)[/tex] by moving all terms involving [tex]\(x\)[/tex] to one side and the constant terms to the other side:
[tex]\[ 12x + 12 = 13x - 12 \][/tex]
Subtract [tex]\(12x\)[/tex] from both sides:
[tex]\[ 12 = x - 12 \][/tex]
Add 12 to both sides:
[tex]\[ 24 = x \][/tex]
Therefore, equation (A) has exactly one solution, [tex]\(x = 24\)[/tex].
### Equation (B)
[tex]\[ -13x + 12 = 13x + 13 \][/tex]
Again, we'll isolate [tex]\(x\)[/tex] by moving all terms involving [tex]\(x\)[/tex] to one side and the constant terms to the other side:
[tex]\[ -13x + 12 = 13x + 13 \][/tex]
Add [tex]\(13x\)[/tex] to both sides:
[tex]\[ 12 = 26x + 13 \][/tex]
Subtract 13 from both sides:
[tex]\[ -1 = 26x \][/tex]
Divide both sides by 26:
[tex]\[ x = -\frac{1}{26} \][/tex]
Therefore, equation (B) has exactly one solution, [tex]\(x = -\frac{1}{26}\)[/tex].
### Equation (C)
[tex]\[ -13x + 12 = 13x - 13 \][/tex]
Isolating [tex]\(x\)[/tex], move all terms involving [tex]\(x\)[/tex] to one side and the constants to the other side:
[tex]\[ -13x + 12 = 13x - 13 \][/tex]
Add [tex]\(13x\)[/tex] to both sides:
[tex]\[ 12 = 26x - 13 \][/tex]
Add 13 to both sides:
[tex]\[ 25 = 26x \][/tex]
Divide both sides by 26:
[tex]\[ x = \frac{25}{26} \][/tex]
Therefore, equation (C) has exactly one solution, [tex]\(x = \frac{25}{26}\)[/tex].
### Equation (D)
[tex]\[ 12x + 12 = 13x + 12 \][/tex]
Isolating [tex]\(x\)[/tex], move all terms involving [tex]\(x\)[/tex] to one side and the constants to the other side:
[tex]\[ 12x + 12 = 13x + 12 \][/tex]
Subtract [tex]\(12x\)[/tex] from both sides:
[tex]\[ 12 = x + 12 \][/tex]
Subtract 12 from both sides:
[tex]\[ 0 = x \][/tex]
Therefore, equation (D) has exactly one solution, [tex]\(x = 0\)[/tex].
### Conclusion
Upon reviewing the calculations, it is evident that each of the equations (A), (B), (C), and (D) has exactly one solution.
So, the correct answer is:
- (A)
- (B)
- (C)
- (D)
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