Question

ребят. срочно. 20 баллов! с решением и объяснением, пожалуйста​

Answer 1

Ответ:на фото

Объяснение:

Answer 2

Ответ:

a)

$\frac{3 - 2x}{x - 2} = - x$

найдём область допустимых значений:

$x - 2 = 0$

$x - 2 + 2 = 0 + 2$

$x = 2$

$\frac{3 - 2x}{x - 2} = - x.x≠2$

$(x - 2) \times \frac{3 - 2x}{x - 2} - (x - 2)x = - x$

сокращаем на общий делитель x-2:

$3 - 2x = - (x - 2)x$

$3 - 2x + (x - 2)x = 0$

$3 - 2x + {x}^{2} - 2x = 0$

$3 - 4x + {x}^{2} = 0$

${x}^{2} - 4x + 3 = 0$

${x}^{2} - x - 3x + 3 = 0$

$x(x - 1) - 3(x - 1) = 0$

$(x - 1)(x - 3) = 0$

$x - 1 = 0 \\ x - 3 = 0$

$x - 1 = 0 \\ x - 1 + 1 = 0 + 1 \\ x = 1$

$x - 3 = 0 \\ x - 3 + 3 = 0 + 3 \\ x = 3$

$x = 1 \\ x = 3 \\ x≠2$

$x = 1 \\ x = 3$

$x_{1} = 1. x_{2} = 3$

б)

${x}^{4} - 3 {x}^{2} + 2 = 0$

${x}^{2 \times 2} - 3 {x}^{2} + 2 = 0$

${( {x}^{2} )}^{2} - 3 {x}^{2} + 2 = 0$

${t}^{2} - 3t + 2 = 0$

${t}^{2} - t - 2t + 2 = 0$

$t(t - 1) - 2(t - 1) = 0$

$(t - 1)(t - 2) = 0$

$t - 1 = 0 \\ t - 2 = 0$

$t = 1 \\ t = 2$

${x}^{2} = 1 \\ {x}^{2} = 2$

${x}^{2} = 1 \\ x = ±1 \\ x = - 1 \\ x = 1$

${x}^{2} = 2 \\ x = ± \sqrt{2} \\ x = - \sqrt{2} \\ x = \sqrt{2}$

$x = - 1 \\ x = 1 \\ x = - \sqrt{2} \\ x = \sqrt{2}$

$x_{1} = - \sqrt{2} . x_{2} = - 1. x_{3} = 1. x_{4} = \sqrt{2}$