Alguem me explica como fazer esses exercicios?
(Raiz de 7)^3
(Raiz quadratica de 9)^2
9x^2 -3x =-5/36
9x² - 3x = - 5/36
9.[x² - (3/9).x] = - 5/36
x² - (3/9).x = - 5/324
x² - (1/3).x = - 5/324
x² - (1/3).x + (1/6)² = - (5/324) + (1/6)²
x² - (1/3).x + (1/6)² = - (5/324) + (1/36)
x² - (1/3).x + (1/6)² = - (5/324) + (9/324)
x² - (1/3).x + (1/6)² = 4/324
x² - (1/3).x + (1/6)² = 1/81
[x - (1/6)]² = [± (1/9)]²
x - (1/6) = ± (1/9)
x = (1/6) ± (1/9)
x = (3/18) ± (2/18)
x = (3 ± 2)/18
x₁ = (3 + 2)/18 → x₁ = 5/18
x₂ = (3 - 2)/18 → x₂ = 1/18
........|¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨ \
........|..........2 Pontos ...... |||"|""\__
........|__________________|||_|___|)
........!(@)';(@)";"""**!(@)(@)****!(@)
Comments
9x² - 3x = - 5/36
9.[x² - (3/9).x] = - 5/36
x² - (3/9).x = - 5/324
x² - (1/3).x = - 5/324
x² - (1/3).x + (1/6)² = - (5/324) + (1/6)²
x² - (1/3).x + (1/6)² = - (5/324) + (1/36)
x² - (1/3).x + (1/6)² = - (5/324) + (9/324)
x² - (1/3).x + (1/6)² = 4/324
x² - (1/3).x + (1/6)² = 1/81
[x - (1/6)]² = [± (1/9)]²
x - (1/6) = ± (1/9)
x = (1/6) ± (1/9)
x = (3/18) ± (2/18)
x = (3 ± 2)/18
x₁ = (3 + 2)/18 → x₁ = 5/18
x₂ = (3 - 2)/18 → x₂ = 1/18
........|¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨ \
........|..........2 Pontos ...... |||"|""\__
........|__________________|||_|___|)
........!(@)';(@)";"""**!(@)(@)****!(@)