Question

1) 5x²-7x+2=0; 2) 2x²+x-6= 0;

3) 4x²-11x-3=0;

4) 6x² +7x+2=0;

5)-2x²-11x+6=0;

Розв'яжіть рівняння:

1) (x-3)(x+3)=2(x+3);

2) (2x-1)² = x(2x-1);

3) (3x-2)²-4x(x+4)=

4;

4) (2x+3)(x-4)-(x + 2)² = -9x.

6) -5x²+4x-19 = 0;

7) x²-8x+2=0;

8) 4+6x-x² = 0;

9) 2x² +8x = 1;

10) 0,16x²-0,8x+9=8.​

Answer 1

Объяснение:

не мало ли для большого количества заданий? Сделаю только первые пять примеров

1)

$5 {x}^{2} - 7x + 2 = 0 \\ d = (7 {x)}^{2} - 4 \times 5 \times 2 = 49 - 40 = 9 \\ x1 = \frac{7 + 3}{10} = \frac{10}{10} = 1 \\ x2 = \frac{7 - 3}{10} = \frac{4}{10} = \frac{2}{5} = 0.4$

2)

$2 {x}^{2} + x - 6 = 0 \\ d = (1 {)}^{2} - 4 \times 2 \times ( - 6) = 1 + 48 = 49 \\ x1 = \frac{ - 1 + 7}{4} = \frac{6}{4} = \frac{3}{2} = 1.5 \\ x2 = \frac{ - 1 - 7}{4} = - \frac{8}{4} = - 2$

3)

$4 {x}^{2} - 11x - 3 = 0 \\ d = (11 {)}^{2} - 4 \times 4 \times ( - 3) = 121 + 48 = 169 \\ x1 = \frac{11 + 13}{8} = \frac{24}{8} = 3 \\ x2 = \frac{11 - 13}{8} = - \frac{2}{8} = - \frac{1}{4} = 0.25$

4)

$6 {x}^{2} + 7x + 2 = 0 \\ d = (7 {)}^{2} - 4 \times 6 \times 2 = 49 - 48 = 1 \\ x1 = \frac{ - 7 + 1}{12} = - \frac{6}{12} = - \frac{1}{6} \\ x2 = \frac{ - 7 - 1}{12} = - \frac{8}{12} = - \frac{2}{3}$

5)

$- 2 {x}^{2} - 11x + 6 = 0 \\ d = (11 {)}^{2} - 4 \times ( - 2) \times 6 = 121 + 48 = 169 \\ x1 = \frac{11 + 13}{ - 4} = \frac{24}{ - 4} = - 6 \\ x2 = \frac{11 - 13}{ - 4} = \frac{ - 2}{ - 4} = \frac{1}{2} = 0.5$