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1 = x(x-2)=2(x+6)
2= x(2x-1)+6=4(x+1)
3= (x-1) (x-2)=6
Soluções para a tarefa
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1
x(2x-1)+6 = 4(x+1)
2x² - x + 6 = 4x + 4
2x² - x - 4x + 6 - 4 = 0
2x² - 5x + 2 = 0
∆ = b² - 4ac
∆ = (-5)² - 4(2)(2)
∆ = 25 - 16
∆ = 9
x = -b ± √∆ /2a
x = -(-5) ± √9 /2.1
x = 5 ± 3 /2
x' = 5+3 /2 = 4
x'' = 5-3 /2 = 1
S = { 4;1 }
(3x+1)²+(x-2)(x+1) = -1
((3x)² + 2(3x)(1) + (1)²) + x² + x - 2x - 2 = -1
(9x² + 6x + 1) + x² - x -2 + 1 = 0
10x² + 5x = 0
10x² + 5x = 0
5x(2x + 1) = 0
5x = 0
x = 0/5
• x = 0
ou
2x + 1 = 0
2x = -1
• x = -1/2
S = { 0;-1/2 }
2x² - x + 6 = 4x + 4
2x² - x - 4x + 6 - 4 = 0
2x² - 5x + 2 = 0
∆ = b² - 4ac
∆ = (-5)² - 4(2)(2)
∆ = 25 - 16
∆ = 9
x = -b ± √∆ /2a
x = -(-5) ± √9 /2.1
x = 5 ± 3 /2
x' = 5+3 /2 = 4
x'' = 5-3 /2 = 1
S = { 4;1 }
(3x+1)²+(x-2)(x+1) = -1
((3x)² + 2(3x)(1) + (1)²) + x² + x - 2x - 2 = -1
(9x² + 6x + 1) + x² - x -2 + 1 = 0
10x² + 5x = 0
10x² + 5x = 0
5x(2x + 1) = 0
5x = 0
x = 0/5
• x = 0
ou
2x + 1 = 0
2x = -1
• x = -1/2
S = { 0;-1/2 }
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