Помогите решить первое и второе задание.

Question 1

Question 2

1)
$\frac{9 {a}^{2} - 16 {b}^{2} }{7a} \times ( \frac{3b - 4a}{4 {b}^{2} - 3ab } - \frac{3b + 4a}{4 {b}^{2} + 3ab } ) = \frac{(3a - 4b)(3a + 4b)}{7a} \times ( \frac{3b - 4a}{b(4b - 3a)} - \frac{3b + 4a}{b(4b + 3a} ) = \frac{(3a - 4b)(3a + 4b)}{7a} \times \frac{(3b - 4a)(4b + 3a) - (3b + 4a)(4b - 3a)}{b(4b - 3a)(4b + 3a)} = \frac{ - (4b - 3a)(4b + 3a)}{7a} \times \frac{12 {b}^{2} - 16ab + 9ab - 12 {a}^{2} - 12 {b}^{2} - 16ab + 9ab + 12 {a}^{2} }{b(4b - 3a)(4b + 3a)} = \frac{ - ( - 14ab)}{7ab} = 2$
Ответ: 2.

2)
$\frac{4x - 6}{x + 2} - \frac{x}{x + 1} = \frac{9}{(x + 1)(x + 2)} \\ \frac{(4x - 6)( x + 1) - x(x + 2)}{(x + 2)(x + 1)} = \frac{9}{(x + 1)(x + 2)} \\ \frac{4 {x}^{2} - 6x + 4x - 6 - {x}^{2} - 2x }{(x + 2)(x + 1)} = \frac{9}{(x + 1)(x + 2)} \\ 3 {x}^{2} - 4x - 6 = 9 \\ 3 {x}^{2} - 4x - 15 = 0 \\ d = {b}^{2} - 4ac = 16 - 4 \times 3 \times ( - 15) = 16 + 180 = 196 \\ x1 = \frac{4 + 14}{2 \times 3} = \frac{18}{6} = 3 \\ x2 = \frac{4 - 14}{2 \times 3} = \frac{ - 10}{6} = - \frac{5}{3}$
Ответ: -5/3; 3.

Question 3

▪1.

$\frac{9 {a}^{2} - 16 {b}^{2} }{7a} \times ( \frac{3b - 4a}{4 {b}^{2} - 3ab} - \frac{3b + 4a}{4 {b}^{2} + 3ab} ) = \frac{(3a - 4b)(3a + 4b) \times (3b - 4a)}{7ab(4b - 3a)} - \frac{(3a - 4b)(3a + 4b) \times (3b + 4a)}{7ab(4b + 3a)} = - \frac{(3a + 4b)(3b - 4a)}{7ab} - \frac{(3a - 4b)(3b + 4a)}{7ab} = - \frac{(3a + 4b)(3b - 4a) + (3a - 4b)(3b + 4a)}{7ab} = - \frac{9ab - 12 {a}^{2} + 12ab - 16ab + 9ab + 12 {a}^{2} - 12 ab - 16ab}{7ab} = - \frac{18ab - 32ab}{7ab} = - \frac{ - 14ab}{7ab} = - ( - 2) = 2$

▪2.

$\frac{4x - 6}{x + 2} - \frac{x}{x + 1} = \frac{9}{(x + 1)(x + 2)} \\ \frac{(4x - 6)(x + 1) - x(x + 2)}{(x + 1)(x + 2)} = \frac{9}{(x + 1)(x + 2)} \: \: \: \: | \times (x + 1)(x + 2) \\ 4 {x}^{2} + 4x - 6x - 6 - {x}^{2} - 2x = 9 \\ 3 {x}^{2} - 4x - 15 = 0 \\ d = 16 + 180 = 196 \\ x1 = \frac{4 + \sqrt{196} }{6} = \frac{4 + 14}{6} = 3 \\ x2 = \frac{4 - 14}{6} = - \frac{5}{3} = - 1 \frac{2}{3}$