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August 2022
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найдите область определения функции
________
y= √ 12-4x-x^2
---------------
1-x
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Universalka
Verified answer
1) 12 - 4x - x² ≥ 0 2) 1 - x ≠ 0
x² + 4x - 12 ≤ 0 x ≠ 1
(x + 6)(x - 2) ≤ 0
+ -- +
____________₀____________
- 6 1 2
x ∈ [- 6 ; 1)∪(1 ; 2]
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Answers & Comments
Verified answer
1) 12 - 4x - x² ≥ 0 2) 1 - x ≠ 0x² + 4x - 12 ≤ 0 x ≠ 1
(x + 6)(x - 2) ≤ 0
+ -- +
____________₀____________
- 6 1 2
x ∈ [- 6 ; 1)∪(1 ; 2]