Question

Помогите мне пожалуйста ​

Answer 1

Объяснение:

1)

$3 {x}^{2} - 4x = 0 \\ x(3x - 4) = 0 \\ 1)x = 0 \\ 2)3x - 4 = 0 \\ 3x = 4 \\ x = \frac{4}{3} \\ x = 1 \frac{1}{3}$

2)

$4 {x}^{2} - 9 = 0 \\ 4 {x}^{2} = 9 \\ {x}^{2} = \frac{9}{4} \\ x = + - \frac{3}{2} = + - 1 \frac{1}{2} = + - 1.5$

3)

$- 5 {x}^{2} + 6x = 0 \\ x( - 5x + 6) = 0 \\ 1)x = 0 \\ 2) - 5x + 6 = 0 \\ - 5x = - 6 \\ x = \frac{6}{5} = 1 \frac{1}{5} = 1.2$

4)

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

5)

${y}^{2} - \frac{1}{9} = 0 \\ {y}^{2} = \frac{1}{9} \\ y = + - \frac{1}{3}$

6)

$6 {u}^{2} - u = 0 \\ u(6u - 1) = 0 \\ 1)u = 0 \\ 2)6u - 1 = 0 \\ 6u = 1 \\ u = \frac{1}{6}$

7)

$6 {m}^{2} - 1 = 0 \\ 6 {m}^{2} = 1 \\ {m}^{2} = \frac{1}{6} \\ m = + - \frac{1}{ \sqrt{6} } = + - \frac{ \sqrt{6} }{6}$

8)

$2y + {y}^{2} = 0 \\ y(2 + y) = 0 \\ 1)y = 0 \\ 2)2 + y = 0 \\ y = - 2$