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Catolker
@Catolker
July 2022
2
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[tex] \sqrt{-2x+3} =x[/tex]
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ShirokovP
Verified answer
ОДЗ
x ≥ 0
- 2x + 3 = x^2
x^2 + 2x - 3 = 0
D= 4 + 12 = 16
x1 = ( - 2 + 4)/2 = 1
x2 = ( - 2 - 4)/2 = - 3 не удовлетворяет ОДЗ
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233334
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Answers & Comments
Verified answer
ОДЗx ≥ 0
- 2x + 3 = x^2
x^2 + 2x - 3 = 0
D= 4 + 12 = 16
x1 = ( - 2 + 4)/2 = 1
x2 = ( - 2 - 4)/2 = - 3 не удовлетворяет ОДЗ
Ответ
1
Ответ:1.