Question

Помогите решить
Алгебра 8 класс ​

Answer 1

Ответ:

1.

А)

а = 3

b = - 5

c = 6

Б)

a = 1

b = - 4

c = 0

В)

a = - 1

b = 0

c = - 7

Г)

a = 1.8

b = 0

c = 0

2.

А)

$9 {x}^{2} - 8x + 4 = 0$

Б)

$- {x}^{2} + 0.3 = 0$

В)

$15 {x}^{2} - 2x = 0$

3.

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

$- 4 {x}^{2} - 4 = 0 \\ - 4 {x}^{2} = 4 \\ {x}^{2} = - 1 \\ {x}^{2} + 1 = 0$

Уравнение не имеет действительных корней

$- 7 {x}^{2} + 14 = 0 \\ - 7 {x}^{2} = - 14 \\ {x}^{2} = - 14 \div ( - 7) \\ {x}^{2} = 2 \\ {x}^{2} - 2 = 0 \\ (x - \sqrt{2} )(x + \sqrt{2} ) = 0 \\ 1)x - \sqrt{2} = 0 \\ x = \sqrt{2} \\ 2)x + \sqrt{2} = 0 \\ x = - \sqrt{2}$

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

$25 {x}^{2} - 9 = 0 \\ 25 {x}^{2} = 9 \\ {x}^{2} = \frac{9}{25} \\ {x}^{2} - \frac{9}{25} = 0 \\ (x - \frac{3}{5} )(x + \frac{3}{5} ) = 0 \\ 1)x - \frac{3}{5} = 0 \\ x = \frac{3}{5} \\ 2)x + \frac{3}{5} = 0 \\ x = - \frac{3}{5}$

$- 2 {x}^{2} + 7x = 0 \\ - x(2x - 7) = 0 \\ 1) - x = 0 \\ x = 0 \\ 2)2x - 7 = 0 \\ 2x = 7 \\ x = 7 \div 2 \\ x = 3.5$

$10 {x}^{2} = 0 \\ {x}^{2} = 0 \\ x = 0$

$0.3 {x}^{2} + 8x = 0 \\ x(0.3x + 8) = 0 \\ 1)x = 0 \\ 2)0.3x + 8 = 0 \\ 0.3x = - 8 \\ x = - 8 \div 0.3 \\ x = - 8 \div \frac{3}{10} \\ x = - 8 \times \frac{10}{3} \\ x = - \frac{80}{3} \\ x = - 26 \frac{2}{3}$

4.

$10 - {(x + 3)}^{2} = x(6 - x) + 4 \\ 10 - ({x}^{2} + 6x + 9) = - x - {x}^{2} + 4 \\ 10 - {x}^{2} - 6x - 9 + x + {x}^{2} - 4 = 0 \\ - 5x - 5 = 0 \\ - 5x = 5 \\ x = - 1$

${(2x + 4)}^{2} = 10(x + 1.6) \\ 4 {x}^{2} + 16x + 16 = 10x + 16 \\ 4 {x}^{2} + 16x + 16 - 10x - 16 = 0 \\ 4 {x}^{2} + 6x = 0 \\ 2x(2x + 3) = 0 \\ 1)2x = 0 \\ x = 0 \\ 2)2x + 3 = 0 \\ 2x = - 3 \\ x = - 3 \div 2 \\ x = - 1.5$

$(3x - 1)(x - 4) = 2x(x - 6.5) \\ 3 {x}^{2} - 12x - x + 4 = 2 {x}^{2} - 13x \\ 3 {x}^{2} - 13x + 4 - 2 {x}^{2} + 13x = 0 \\ {x}^{2} + 4 = 0$

Уравнение не имеет действительных корней