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PHRF
@PHRF
April 2020
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Bom dia, Por favor ajuda:
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ittalo25
√(16-x²) = √(12-(x-1)²) + √3
√(16-x²) = √(12-(x²-2x+1)) + √3
√(16-x²) = √(12-x²+2x-1) + √3
(√(16-x²))² = (√(12-x²+2x-1) + √3)²
16-x² = 12-x²+2x - 1 + (2√3 .(12-x²+2x-1)) + 3
16 = 12 + 2x -1 + (√12 . (12-x²+2x-1)) + 3
16 = 12 + 2x -1 + 12√12 - x²√12 + x2√12 -√12 + 3
16 -12 +1 -3 = 2x + 12√12 - x²√12 + x2√12 -√12
2 = 2x + 12√12 - x²√12 + x2√12 -√12
2x + 12√12 - x²√12 + x2√12 -√12 -2 = 0
√12x² + x.(2√12+2) + 11√12 -2 = 0
bhaskara:
(2√12+2)² - 4 . √12 . (11√12 -2) = Δ
tenta fazer a partir daqui
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Lista de comentários
√(16-x²) = √(12-(x²-2x+1)) + √3
√(16-x²) = √(12-x²+2x-1) + √3
(√(16-x²))² = (√(12-x²+2x-1) + √3)²
16-x² = 12-x²+2x - 1 + (2√3 .(12-x²+2x-1)) + 3
16 = 12 + 2x -1 + (√12 . (12-x²+2x-1)) + 3
16 = 12 + 2x -1 + 12√12 - x²√12 + x2√12 -√12 + 3
16 -12 +1 -3 = 2x + 12√12 - x²√12 + x2√12 -√12
2 = 2x + 12√12 - x²√12 + x2√12 -√12
2x + 12√12 - x²√12 + x2√12 -√12 -2 = 0
√12x² + x.(2√12+2) + 11√12 -2 = 0
bhaskara:
(2√12+2)² - 4 . √12 . (11√12 -2) = Δ
tenta fazer a partir daqui