Question

погите решить три уравнения очень срочнооооо​

Answer 1

$1) \frac{ {x}^{2} }{x + 3} = \frac{9}{x + 3} \times x + 3 \\ {x}^{2} - 9 = 0 \\ {x}^{2} = 9$

$x1 = 3 \\ x2 = - 3$

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

$- {x}^{2} + 2x + 3 = 0 \times ( - 1) \\ {x}^{2} - 2x - 3 = 0 \\ d = 4 + 12 = 16 = 4$

$x1 = \frac{2 + 4}{2} = \frac{6}{2} = 3 \\ x2 = \frac{2 - 4}{2} = - \frac{2}{2} = - 1$

3 не знаю