Question

Помогите пожалуйста ​

Answer 1

Ответ:

1

$\frac{3x}{x -1 } - \frac{2x}{x + 2} = \frac{3x - 6}{(x - 1)(x + 2)} \\ \\ x - 1 \ne0 = > x \ne1\\ x + 2 \ne0 = > x\ne - 2 \\ \\ 3x(x + 2) - 2x(x - 1) = 3x - 6 \\ 3 {x}^{2} + 6x - 2 {x}^{2} + 2x = 3x - 6 \\ {x}^{2} + 5x + 6 = 0 \\ D = 25 - 24 = 1 \\ x_1 = \frac{ - 5 + 1}{2} = - 2(\text{не подходит}) \\ x_2 = - 3$

Ответ: -3

2.

$x + 1 = \sqrt{ 7x - 5} \\ \\\text{ОДЗ:} \\ \left \{ {{7x - 5 \geqslant 0} \atop {x + 1 \geqslant 0} } \right. \\ \\ \left \{ {{x \geqslant \frac{5}{7} } \atop {x \geqslant - 1} } \right. \\ = > x \geqslant \frac{5}{7} \\ \\ {(x + 1)}^{2} = 7x - 5 \\ {x}^{2} + 2x + 1 = 7x - 5 \\ {x}^{2} - 5x + 6 = 0 \\ D= 25 - 24 = 1 \\ x_1 = \frac{5 + 1}{2} = 3 \\ x_2 = 2$

Ответ: 2, 3.

3.

а

${2}^{2x} + 14 \times {2}^{2x + 1} = 29 \\ {2}^{2x} (1 + 14 \times 2) = 29 \\ {2}^{2x} \times 29 = 29 \\ {2}^{2x} = 1 \\ 2x = 0 \\ x = 0$

б

${9}^{x} - 6 \times {3}^{x} - 27 = 0 \\ \\ {3}^{x} = t \\ t {}^{2} - 6t - 27= 0 \\ D = 36 + 108 = 144 \\ t_1 = \frac{6 + 12}{2} = 9 \\ t_2 = - 3( < 0,\text{не подходит} )\\ \\ {3}^{x} = 9 \\ x = 2$

3.

а

$log_{2}(x + 14) = log_{2}(64) - log_{2}(x + 2) \\ \\ \text{ОДЗ:} \\ \left \{ {{x + 14 > 0} \atop {x + 2 > 0} } \right. \\ \\ \left \{ {{x > - 14} \atop {x > - 2} } \right. \\ = > x > - 2 \\ \\ log_{2}(x + 14) + log_{2}(x + 2) = log_{2}(64) \\ log_{2}((x + 14)(x + 2)) = log_{2}(64) \\ {x}^{2} + 14x + 2x + 28 = 64 \\ {x}^{2} + 16x - 36 = 0 \\ D = 256 + 144 = 400 \\ x_1 = \frac{ - 16 + 20}{2} = 2 \\ x_2 = - 18(\text{не подходит})$

Ответ: 2

б

$2log_{3} {}^{2} (x) - log_{3}( {x}^{7} ) + 3 = 0 \\ 2 log_{3} {}^{2} (x) - 7log_{3}(x) + 3 = 0 \\ \\ \text{ОДЗ:} \\ x > 0 \\ \\ log_{3}(x) = t \\2 t {}^{2} - 7t + 3 = 0 \\ D = 4 9- 24 = 25 \\ t_1 = \frac{7 + 5}{4} = 3 \\ t_2 = \frac{1}{2} \\ \\ log_{3}(x) = 3 \\ x1 = 27 \\ \\ log_{3}(x) = 0.5 \\ x2 = \sqrt{3}$

4.

а

$8 \sin {}^{2} (x) - 2 \cos(x) = 5 \\ 8 - 8 \cos {}^{2} (x) - 2 \cos(x) - 5 = 0 \\ 8 \cos {}^{2} (x) + 2 \cos(x) - 3 = 0 \\ \\ \cos(x) = t \\ 8t {}^{2} + 2t - 3 = 0 \\ D = 4 + 96 = 100 \\ t_1 = \frac{ - 2 + 10}{16} = \frac{1}{2} \\ t_2 = - \frac{12}{16} = - \frac{3}{4} \\ \\ \cos(x) = \frac{1}{2} \\ x_1 = \pm \frac{\pi}{3} + 2\pi \: n \\ \\ \cos(x) = - \frac{3}{4} \\ x_2 = \pi\pm \: arccos( \frac{3}{4} ) + 2\pi \: n$

n принадлежит Z.

б

$2 \sin(x) \cos(x ) + 3 \cos {}^{2} (x) = \sin {}^{2} (x) \\ \sin {}^{2} (x) - 2 \sin(x) \cos(x) - 3 \cos {}^{2} (x) = 0 \: \: \: | \div \cos {}^{2} (x) \ne0 \\ {tg}^{2} x - 2tgx - 3 = 0 \\ \\ tgx = t \\ t {}^{2} - 2 t - 3 = 0 \\ D = 4 + 12 = 16 \\ t_1 = \frac{2 + 4}{2} = 3 \\ t_2 = - 1 \\ \\ tgx = 3 \\ x_1 = arctg(3) + \pi \: n \\ \\ tgx = - 1 \\ x_2 = - \frac{\pi}{4} + \pi \: n$

n принадлежит Z.