Question

x³-x=0
4y³-4y=0
(x-4) (2x+1)​

Answer 1

${x}^{3} - x = 0 \\ x( {x}^{2} - 1 )= 0 \\ x = 0 < > {x}^{2} - 1 = 0 \\ {x}^{2} = 1 \\ x = 1 \\ x = - 1$

Ответ: х=-1, х=0, х=1.

$4 {y}^{3} - 4y = 0 \\ 4y( {y}^{2} - 1) = 0 \\ 4y = 0 < > {y}^{2} - 1 = 0 \\ y = 0 < > {y}^{2} = 1 \\ y = 1 \\ y = - 1$

Ответ: у=-1, у=0, у=1.

$(x - 4)(2x + 1) = 0 \\ 2 {x}^{2} + x - 8x - 4 =0 \\ 2 {x}^{2} - 7x - 4 = 0 \\ d = 49 - 4 \times 2 \times ( - 4) = 81 = {9}^{2} \\ x = \frac{7 - 9}{2 \times 2} = \frac{ - 2}{4} = - 0.5 \\ x = \frac{7 + 9}{2 \times 2} = \frac{16}{4} = 4$

Ответ: х=0.5, х=4.