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zlata25148
@zlata25148
August 2022
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Решите задание на фото
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IrkaShevko
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ОДЗ: 2x + 5 ≠ 0; x + 4 ≠ 0; 6 + x - x² ≥ 0
x² - x - 6 ≤ 0
x ≠ -2,5
x ≠ -4
D = 1 + 24 = 25
x1 = (1 - 5)/2 = -2
x2 = (1 + 5)/2 = 3
x∈[-2; 3]
1) т.к. на ОДЗ знаменатели положительные, а числители равны, то
2x + 5 ≤ x + 4
x ≤ -1
Ответ: x∈[-2; -1]
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Answers & Comments
Verified answer
ОДЗ: 2x + 5 ≠ 0; x + 4 ≠ 0; 6 + x - x² ≥ 0x² - x - 6 ≤ 0
x ≠ -2,5
x ≠ -4
D = 1 + 24 = 25
x1 = (1 - 5)/2 = -2
x2 = (1 + 5)/2 = 3
x∈[-2; 3]
1) т.к. на ОДЗ знаменатели положительные, а числители равны, то
2x + 5 ≤ x + 4
x ≤ -1
Ответ: x∈[-2; -1]