Question

resolva as seguintes equacoes fracionarias pfv
1)x/2 + x/x + 2=x + 4/4
2)1/x - 3 - 1/2=1/x - 2​

Answer 1

Resposta:

a) x ≠ -2

$\frac{x}{2} + \frac{x}{x + 2} = \frac{x + 4}{4} \\ \frac{2x(x + 2) + 4x}{4(x + 2)} = \frac{(x + 2)(x + 4)}{4(x + 2)} \\ 2x(x + 2) + 4x = (x + 2)(x + 4) \\ 2 {x}^{2} + 4x + 4x = {x}^{2} + 4x + 2x + 8 \\ 2 {x}^{2} - {x}^{2} + 4x + 4x - 4x - 2x - 8 = 0 \\ {x}^{2} + 2x - 8 = 0$

∆ = b² - 4ac

∆ = 2² - 4.1.(-8)

∆ = 4 + 32

∆ = 36

$x = \frac{ - b \pm \sqrt{ \Delta} }{2a} \\ x = \frac{ - 2 \pm \sqrt{36} }{2.1} \\ x = \frac{ - 2 \pm6}{2} \\ x = - 1 \pm3 \\ x' = - 1 + 3 = 2 \\ x'' = - 1 - 3 = - 4$

S = {-4; 2}

b) x ≠ 3 e x ≠ 2

$\frac{1}{x - 3} - \frac{1}{2} = \frac{1}{x - 2} \\ \frac{2(x - 2) - (x - 3)(x - 2)}{2(x - 3)(x - 2)} = \frac{2(x - 3)}{2(x - 3)(x - 2)} \\ 2(x - 2) - (x - 3)(x - 2) = 2(x - 3) \\ 2x - 4 - ( {x}^{2} - 2x - 3x + 6) = 2x - 6 \\ 2x - 4 - {x}^{2} + 2x + 3x - 6 = 2x - 6 \\ - {x}^{2} + 5x - 4 = 0 \\ {x}^{2} - 5x + 4 = 0$

∆ = b² - 4ac

∆ = (-5)² - 4.1.4

∆ = 25 - 16

∆ = 9

$x = \frac{ - b \pm \sqrt{ \Delta} }{2a} \\ x = \frac{ - ( - 5) \pm \sqrt{9} }{2.1} \\ x = \frac{5 \pm3}{2} \\ x' = \frac{5 + 3}{2} = \frac{8}{2} = 4 \\ x'' = \frac{5 - 3}{2} = \frac{2}{2} = 1$

S = {1; 4}