Question

Помогите,с решением только 3

Answer 1

Ответ:

-(х-3)×(х-2)

ну собственно ответ выше

Answer 2

${x}^{2} - 2x - 15 = 0 \\ d = 4 - 4 \times 1 \times ( - 15) = 4 + 60 = 64 = {8}^{2} \\ x1 = \frac{2 + 8}{2} = 5 \\ x2 = \frac{2 - 8}{2} = - 3 \\ {x}^{2} + 3x - 10 = 0 \\ d = 9 - 4 \times 1 \times ( - 10) = 9 + 40 = 49 = {7}^{2} \\ x1 = \frac{ - 3 + 7}{2} = 2 \\ x2 = \frac{ - 3 - 7}{2} = - 5 \\ - {x}^{2} + 5x - 6 \\ {x}^{2} - 5x + 6 = 0 \\ d = 25 - 4 \times 1 \times 6 = 25 - 24 = 1 \\ x1 = \frac{5 + \sqrt{1} }{2} \\ x2 = \frac{5 - \sqrt{1} }{2}$