Question

Помогите пожалуйста с алгеброй 8 класс

Answer 1

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

$\frac{3 - 2x}{x - 2} = - x,x≠2$

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

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

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

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

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

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

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

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

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

х = 1

,х≠2

х = 3

$x = 1 \\ x = 3$

$x _{1} = 1,x _{2}= 3$

_____________________

$б)x {}^{4} - 3x {}^{2} + 2 = 0$

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

$t = 1 \\ t = 2$

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

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

$x _{1} = - \sqrt[]{2} ,x _{2} = - 1,x _{3} = 1,x _{4} = \sqrt{2}$