Question

1) Solucione as equações abaixo:

a) X² - 4x = 0 b) 3x² + 6x = 0 c) 2x² - 7x = 0 d) – x² + 5x = 0 e) (x – 2). ( x – 3) – 6 = 0 f) (2x + 3)² = 9 + 5x g) (x + 2). (2x – 3) = -6 + 5x h) ( x – 3)²- 9 = 0][

Answer 1

Resposta:

$a) \: \: \: \: \: {x}^{2} - 4x = 0 \\ x \times (x - 4) = 0 \\ ou \\ x - 4 = 0 \\ x = 4 \\ \\ b) \: \: \: \: \: {3x}^{2} + 6x = 0 \\ x \times (3x + 6) = 0 \\ ou \\ 3x + 6 = 0 \\ 3x = - 6 \\ x = - \frac{6}{3} \\ x = - 2 \\ \\ c) \: \: \: \: \: {2x}^{2} - 7x = 0 \\ x \times (2x - 7) = 0 \\ ou \\ 2x - 7 = 0 \\ 2x = 7 \\ x = \frac{7}{2} \\ x = 3.5 \\ \\ d) \: \: \: \: \: - {x}^{2} + 5x = 0 \\ x \times ( - x + 5) = 0 \\ ou \\ - x + 5 = 0 \\ - x = - 5 \: \: \: \: \: \times ( - 1) \\ x = 5 \\ \\ e) \: \: \: \: \: (x - 2) \times (x - 3) - 6 = 0\\ {x}^{2} - 3x - 2x + 6 - 6 = 0 \\ {x}^{2} - 5x = 0 \\ x \times (x - 5) = 0 \\ ou \\ x - 5 = 0 \\ x = 5 \\ \\ f) \: \: \: \: \: {(2x + 3)}^{2} = 9 + 5x \\ (2x + 3) \times (2x + 3) = 9 + 5x \\ {4x}^{2} + 6x + 6x + 9 = 9 + 5x \\ {4x}^{2} + 6x + 6x + 9 - 9 - 5x = 0 \\ {4x }^{2} + 6x + 6x - 5x + 9 - 9 = 0 \\ {4x}^{2} + 12x - 5x = 0 \\ {4x}^{2} + 7x = 0 \\ x \times (4x + 7) = 0 \\ ou \\ 4x + 7 = 0 \\ 4x = - 7 \\ x = - \frac{7}{4} \\ x = - 1.75 \\ \\ g) \: \: \: \: \: (x + 2) \times (2x - 3) = - 6 + 5x \\ {2x}^{2} - 3x + 4x - 6 = - 6 + 5x \\ {2x}^{2} - 3x + 4x - 6 + 6 - 5x = 0 \\ {2x}^{2} - 3x + 4x - 5x - 6 + 6 = 0 \\ {2x}^{2} + x - 5x = 0 \\ {2x}^{2} - 4x = 0 \\ x \times (2x - 4) = 0 \\ ou \\ 2x - 4 = 0 \\ 2x = 4 \\ x = \frac{4}{2} \\ x = 2 \\ \\h) \: \: \: \: \: {(x - 3)}^{2} - 9 = 0 \\ (x - 3) \times (x - 3) - 9 = 0 \\ {x}^{2} - 3x - 3x + 9 - 9 = 0 \\ {x}^{2} - 6x = 0 \\ x \times (x - 6) = 0 \\ ou \\ x - 6 = 0 \\ x = 6$