Question

СРОЧНО,ПОЖАЛУЙСТА!!!

Answer 1

1.

$x + 4≠0 \\ x ≠ - 4$

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

Ответ: 0; 4

2.

$x + 3≠ 0 \\ x≠ - 3$

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

Ответ: 4

3.

${x}^{2} - 8x + 12 = 0 \\ D = {( - 8)}^{2} - 4 \times 1 \times 12 = 64 - 48 = 16 \\ \sqrt{D} = \sqrt{16} = 4 \\ x1≠ \frac{8 + 4}{2} ≠ \frac{12}{2}≠ 6 \\ x2 ≠ \frac{8 - 4}{2} ≠ \frac{4}{2 } ≠ 2$

${x}^{2} - 5x - 6 = 0 \\ D = {( - 5)}^{2} - 4 \times 1 \times ( - 6) = 25 + 24 = 49 \\ \sqrt{D} = \sqrt{49} = 7 \\ x1 = \frac{5 + 7}{2} = \frac{12}{2} = 6 \\ x2 = \frac{5 - 7}{2} = \frac{ - 2}{2} = - 1$

Ответ: -1