Question

СРОЧНО!!! Вариант 2 помогите решить пожалуйста СРОЧНО!!!

Answer 1

Ответ:

1.

a)

$x - 64 {x}^{3} = 0 \\ x(1 - 64 {x}^{2} ) = 0 \\ \\ x_1 = 0 \\ \\ 1 - 64 {x}^{2} = 0 \\ 64 {x}^{2} = 1 \\ {x}^{2} = \frac{1}{64} \\ x_2 = \frac{1}{8} \\ x_3 = - \frac{1}{8}$

б)

$25 {x}^{2} - {x}^{4} = 0 \\ {x}^{2} (25 - {x}^{2} ) = 0 \\ \\ x_1 = 0 \\ \\ 25 - {x}^{2} = 0 \\ {x}^{2} = 25 \\ x_2 = 5 \\ x_3 = - 5$

2.

${x}^{2} - 17x + 16 = 0 \\ D= 289 - 64 = 225 \\ x_1 = \frac{17 + 15}{2} = 16 \\ x_2 = 1$

3.

а)

${( {x}^{2} + 2)}^{2} - 5( {x}^{2} + 2) + 6 = 0 \\ \\ {x}^{2} + 2 = t \\ \\ t {}^{2} - 5t + 6 = 0 \\ D = 25 - 24 = 1\\ t_1 = \frac{5 + 1}{2} = 3 \\ t_2 = 2 \\ \\ {x}^{2} + 2 = 3 \\ {x}^{2} = 1 \\ x_{1,2} = \pm1 \\ \\ {x}^{2} + 2 = 2 \\ {x}^{2} = 0 \\ x_3 = 0$

б)

$(x - (3x + 6))(x - (3x - 6)) = - 32 \\ (x - 3x - 6)(x - 3x + 6) = - 32 \\ ( - 2x - 6)( - 2x + 6) = - 32 \\ {( - 2x)}^{2} - {6}^{2} = - 32 \\ 4 {x}^{2} - 36 = - 32 \\ 4 {x}^{2} = 4 \\ {x}^{2} = 1 \\ x_1 = 1 \\ x_2 = - 1$

в)

$\frac{ {x}^{2} - x}{ {x}^{2} - x + 1} - \frac{ {x}^{2} - x + 2 }{ {x}^{2} - x - 2 } = 1 \\ \\ {x}^{2} - x = t \\ \\ \frac{t}{ t+ 1} - \frac{t + 2}{t - 2} = 1 \\ \\ t\ne - 1 \\ t\ne2 \\ \\ t(t - 2) - (t + 2)(t + 1) = (t + 1)(t - 2) \\ {t}^{2} - 2t - ( {t}^{2} + 2 t + t + 2) = {t}^{2} + t - 2t - 2 \\ {t}^{2} - 2 t - {t}^{2} - 3t - 2 = {t}^{2} - t - 2 \\ {t}^{2} + 4t = 0 \\ t(t + 4) = 0 \\ t_1 = 0 \\ t_2 = - 4 \\ \\ {x}^{2} - x = 0 \\ x(x - 1) = 0 \\ x_1 = 0 \\ x_2 = 1 \\ \\ {x}^{2} - x = - 4 \\ {x}^{2} - x + 4 = 0$

нет корней

Ответ: 0; 1