Ответ:на фото

Объяснение:

Verified answer

Ответ:

a)

[tex] \frac{3 - 2x}{x - 2} = - x[/tex]

найдём область допустимых значений:

[tex]x - 2 = 0[/tex]

[tex]x - 2 + 2 = 0 + 2[/tex]

[tex]x = 2[/tex]

[tex] \frac{3 - 2x}{x - 2} = - x.x≠2[/tex]

[tex](x - 2) \times \frac{3 - 2x}{x - 2} - (x - 2)x = - x[/tex]

сокращаем на общий делитель x-2:

[tex]3 - 2x = - (x - 2)x[/tex]

[tex]3 - 2x + (x - 2)x = 0[/tex]

[tex]3 - 2x + {x}^{2} - 2x = 0[/tex]

[tex]3 - 4x + {x}^{2} = 0[/tex]

[tex] {x}^{2} - 4x + 3 = 0[/tex]

[tex] {x}^{2} - x - 3x + 3 = 0[/tex]

[tex]x(x - 1) - 3(x - 1) = 0[/tex]

[tex](x - 1)(x - 3) = 0[/tex]

[tex]x - 1 = 0 \\ x - 3 = 0[/tex]

[tex]x - 1 = 0 \\ x - 1 + 1 = 0 + 1 \\ x = 1[/tex]

[tex]x - 3 = 0 \\ x - 3 + 3 = 0 + 3 \\ x = 3[/tex]

[tex]x = 1 \\ x = 3 \\ x≠2[/tex]

[tex]x = 1 \\ x = 3[/tex]

[tex] x_{1} = 1. x_{2} = 3[/tex]

б)

[tex] {x}^{4} - 3 {x}^{2} + 2 = 0[/tex]

[tex] {x}^{2 \times 2} - 3 {x}^{2} + 2 = 0[/tex]

[tex] {( {x}^{2} )}^{2} - 3 {x}^{2} + 2 = 0[/tex]

[tex] {t}^{2} - 3t + 2 = 0[/tex]

[tex] {t}^{2} - t - 2t + 2 = 0[/tex]

[tex]t(t - 1) - 2(t - 1) = 0[/tex]

[tex](t - 1)(t - 2) = 0[/tex]

[tex]t - 1 = 0 \\ t - 2 = 0[/tex]

[tex]t = 1 \\ t = 2[/tex]

[tex] {x}^{2} = 1 \\ {x}^{2} = 2[/tex]

[tex] {x}^{2} = 1 \\ x = ±1 \\ x = - 1 \\ x = 1[/tex]

[tex] {x}^{2} = 2 \\ x = ± \sqrt{2} \\ x = - \sqrt{2} \\ x = \sqrt{2} [/tex]

[tex]x = - 1 \\ x = 1 \\ x = - \sqrt{2} \\ x = \sqrt{2} [/tex]

[tex] x_{1} = - \sqrt{2} . x_{2} = - 1. x_{3} = 1. x_{4} = \sqrt{2} [/tex]