Question

Даю 25 баллов, 7.9 задание. Срочно!​

Answer 1

Ответ:

$( \sqrt{(3x - 1)(4x + 3} ) {}^{2} = (3x - 1) {}^{2} \\ (3x - 1)(4x + 3) = 9 {x}^{2} - 6x + 1 \\ 12 {x}^{2} + 9x - 4x - 3 = {9x}^{2} - 6x + 1$

$12 {x}^{2} + 5x - 3 = 9 {x}^{2} - 6x + 1 \\ 12 {x}^{2} + 5x - 3 - 9 {x}^{2} + 6x - 1 = 0 \\ 3 {x}^{2} + 11x - 4 = 0 \\ d = ( {11)}^{2} + 4 \times 3 \times 4 = \\ = 121 + 48 = 169 = {13}^{2} \\ x1 = \frac{ - 11 + 13}{6} = \frac{2}{6} = \frac{1}{3}$

$x2 = \frac{ - 11 - 13}{6} = - \frac{24}{6} = - 4$