Urgente!
Resolver as equações abaixo:
| x – 1 | = 5
| x + 1 | = | 2x – 3 |
| 2x + 1 | = 5x – 2
| x |² - 2 |x | - 3 = 0
Soluções para a tarefa
Resposta:
| x – 1 | = 5
|x-1|=5
x=5-1
x=4
| x + 1 | = | 2x – 3 |
x+2x=-3+1
3x=-2
x=-2/3
| 2x + 1 | = 5x – 2
2x-5x=-2+1
-3x=-1 (-1)
x=1/3
| x |² - 2 |x | - 3 = 0
x^2+2x-3=0
espero que isso ajude vc
Oie, Td Bom?!
1.
|x - 1| = 5
• x - 1 = 5
x = 5 + 1
x₂ = 6
• x - 1 = - 5
x = - 5 + 1
x₁ = - 4
S = {- 4 , 6}
2.
|x + 1| = |2x - 3|
• x + 1 = 2x - 3
x - 2x = - 3 - 1
- x = - 4 . (- 1)
x₂ = 4
• x + 1 = - (2x - 3)
x + 1 = - 2x + 3
x + 2x = 3 - 1
3x = 2
x₁ = 2/3
S = {2/3 , 4}
3.
|2x + 1| = 5x - 2
|2x + 1| - 5x = - 2
• 2x + 1 - 5x = - 2 , 2x + 1 ≥ 0
- 3x + 1 = - 2 , 2x ≥ - 1
- 3x = - 2 - 1 , x ≥ - 1/2
- 3x = - 3
x = -3/-3
x = 3/3
x = 1
• - (2x + 1) - 5x = - 2 , 2x + 1 < 0
- 2x - 1 - 5x = - 2 , 2x < - 1
- 7x - 1 = - 2 , x < - 1/2
- 7x = - 2 + 1
- 7x = - 1
x = -1/-7
x = 1/7
x = 1 , x ≥ - 1/2 ⇒ x = 1
x = 1/7 , x < - 1/2 ⇒ x∈∅
S = {1}
4.
|x|² - 2 . |x| - 3 = 0
x² - 2 . |x| - 3 = 0
• x² - 2x - 3 = 0 , x ≥ 0
x² + x - 3x - 3 = 0
x . (x + 1) - 3(x + 1) = 0
(x + 1) . (x - 3) = 0
x + 1 = 0 ⟹ x₁ = - 1
x - 3 = 0 ⟹ x₂ = 3
• x² - 2 . (- x) - 3 = 0 , x < 0
x² + 2x - 3 = 0
x² + 3x - x - 3 = 0
x . (x + 3) - (x + 3) = 0
(x + 3) . (x - 1) = 0
x + 3 = 0 ⟹ x₁ = - 3
x - 1 = 0 ⟹ x₂ = 1
x₁ = - 1 , x ≥ 0 ⇒ x = 3
x₂ = 3
x₁ = - 3 , x < 0 ⇒ x = - 3
x₂ = 1
S = {- 3 , 3}
Att. Makaveli1996