Question

Пожалуйста плиз срочно​

Answer 1

Ответ:

Объяснение:

1) x(x-14)=0  x=0 x= 14

2)x(12x + 1)= 0  x= 0 x = -1/12

3) x(2x - 2.4) = 0  x =0 x= 1.2

4) y(0.4y - 4) = 0   y = 0 y = 10

5)y^2(1-18y) = 0  y=0 y =1/18

6)y^4(4y-5) = 0 y=0 y =1.25

Answer 2

$1)x {}^{2} - 14x = 0$

$x \times (x - 14) = 0$

$x = 0$

$x - 14 = 0$

$x = 0$

$x = 14$

$x_{1} = 0,x _{2} = 14$

$2)12x {}^{2} + x = 0$

$x \times (12x + 1) = 0$

$x = 0$

$12x + 1 = 0$

$x = 0$

$x = - \frac{1}{12}$

$x _{1} = - \frac{1}{12} ,x _{2} = 0$

$3)2x {}^{2} - 2.4x = 0$

$2x {}^{2} - \frac{12}{5} x = 0$

$10x {}^{2} - 12x = 0$

$2x \times (5x - 6) = 0$

$x \times (5x -6 ) = 0$

$x = 0$

$5x - 6 = 0$

$x = 0$

$x = \frac{5}{6}$

$x _{1} = 0,x _{2} = 1.2$

$4)0.4y {}^{2} - 4y = 0$

$y {}^{2} - 10y = 0$

$y \times (y - 10) = 0$

$y = 0$

$y - 10 = 0$

$y = 0$

$y = 10$

$y _{1} = 0,y _{2} = 10$

$5) \frac{1}{18} y {}^{2} - y {}^{3} = 0$

$y {}^{2} \times ( \frac{1}{18} - y) = 0$

$y {}^{2} = 0$

$y = \frac{1}{18}$

$y _{1} = 0,y _{2} = \frac{1}{18}$