Question

X³-x=04y³-4y=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.

Answer 2

It seems like you've provided an expression, but it's a bit unclear what you're asking or trying to convey. It looks like you have a mixture of equations and expressions involving variables. Could you please provide more context or clarify your question? Are you trying to solve for certain variables, simplify the expression, or understand something specific about this equation? Please provide more information so I can assist you accurately.