Предмет: Алгебра, автор: svvroma

Поооожалуйста помогите алгебра отдал все балы

Приложения:

Ответы

Автор ответа: Miroslava227

Ответ:

${x}^{4} - 6 {x}^{2} - 27 = 0 \\ \\ {x}^{2} = t \\ \\ t {}^{2} - 6 t - 27 = 0 \\D = 36 + 108 = 144\\ t_1 = \frac{6 + 12}{2} = 9 \\ t_2 = - 3 \\ \\ {x}^{2} = 9 \\ x = \pm 3 \\ \\ {x}^{2} = - 3$

нет корней

Ответ: -3; 3.

$2 {x}^{4} - 9 {x}^{2} + 4 = 0 \\ \\ {x}^{2} = t \\ \\2 t {}^{2} - 9t + 4 = 0\\ D = 81 -32 = 49 \\ t_1 = \frac{9 + 7}{4} = 4 \\ t_2 = \frac{1}{2} \\ \\ {x}^{2} = 4 \\ x_{1,2} = \pm2 \\ \\ {x}^{2} = \frac{1}{2} \\ x_{3,4} = \pm \frac{ \sqrt{2} }{2}$

$x - 3 \sqrt{x} - 10 = 0 \\ \\ \sqrt{x} = t \\ t \geqslant 0 \\ \\ t {}^{2} - 3t - 10 = 0\\ D = 9 + 40 = 49\\ t_1 = \frac{3 + 7}{2} = 5 \\ t_2 = - 2 \\ \\ \sqrt{x} = 5 \\ x = 25$

Ответ: 25

$(x - 1) {}^{4} - {(x - 1)}^{2} - 12 = 0 \\ \\ {(x - 1)}^{2} = t \\ t > 0\\ \\ t {}^{2} - t - 12 = 0\\ D= 1 + 48 = 49 \\ t_1 = \frac{1 + 7}{2} = 4 \\ t_2 = - 3 \\ \\ {(x - 1)}^{2} = 4 \\ {(x - 1)}^{2} - 4 = 0 \\ (x - 1 - 2)(x - 1 + 2) = 0 \\ (x - 3)(x + 1) = 0 \\ x_1 = 3 \\ x_2 = - 1$

Ответ: -1; 3

${( {x}^{2} + 3x) }^{2} - 2( {x}^{2} + 3x) - 8 = 0 \\ \\ {x}^{2} + 3x = t \\ \\ t { }^{2} - 2 t - 8 = 0 \\ D = 4 + 32 = 36 \\ t_1 = \frac{2 + 6}{2} = 4 \\ t_2 = - 2 \\ \\ {x}^{2} + 3x = 4 \\ {x}^{2} + 3x - 4 = 0 \\ D = 9 + 16 = 25 \\ x_1 = \frac{ - 3 + 5}{2} = 1 \\ x_2 = - 4\\ \\ {x}^{2} + 3x = - 2 \\ {x}^{2} + 3x + 2 = 0 \\ D = 9 - 8 = 1 \\ x_1 = \frac{ - 3 + 1}{2} = - 1 \\ x_2 = - 2$

Ответ: -4; -2; -1; 1

$(2 {x}^{2} + x + 1)(2 {x}^{2} + x + 3) = 8 \\ (2 {x}^{2} + x + 1)(2 {x}^{2} + x + 1 + 2) = 8 \\ \\ 2 {x}^{2} + x + 1 = t \\ \\ t(t + 2 )= 8 \\ {t}^{2} + 2t - 8 = 0 \\ D= 4 + 32 = 36 \\ t_1 = \frac{ - 2 + 6}{2} = 2 \\ t_2 = - 4 \\ \\ 2 {x}^{2} + x + 1 = 2 \\ 2 {x}^{2} + x - 1 = 0 \\ D = 1 + 8 = 9 \\ x_1 = \frac{ - 1 + 3}{4} = 0.5 \\ {x}^{2} = - 1 \\ \\ 2 {x}^{2} + x + 1 = - 4 \\ 2 {x}^{2} + x + 5 = 0 \\ D = 1 - 4 < 0$