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Verified answer
No. 6syarat daerah asal
6 + x - x² ≥ 0
x² - x - 6 ≤ 0
(x + 2)(x - 3) ≤ 0
-2 ≤ x ≤ 3
syarat pertidaksamaan
√(6 + x - x²) < 2
6 + x - x² < 4
2 + x - x² < 0
x² - x - 2 > 0
(x + 1)(x - 2) > 0
x < -1 atau x > 2
irisan kedua syarat
-2 ≤ x < -1 atau 2 < x ≤ 3