x³-x = 0 4y³-4y = 0 (x-4) (2x+1)

Question

Математика

Вадим

5 год назад

x³-x = 0 4y³-4y = 0 (x-4) (2x+1)

ОТВЕТЫ

Прокл

Oct 5, 2020

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

х=-1, х=0, х=1.

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

у=-1, у=0, у=1.

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

х=0.5, х=4.

824

x³-x = 0 4y³-4y = 0 (x-4) (2x+1)​

x³-x = 0 4y³-4y = 0 (x-4) (2x+1)