Question

Помогите пожалуйста завтра экзамен!!!!!!!!!!!

Answer 1

Решение задания приложено

Answer 2

1.11. По теореме синусов:
AB/sinC = 2R,
2R = 3sqrt(2) : sqrt(2)/2 = 3sqrt(2) * 2/sqrt(2) = 6
R = 3 (см).
Ответ: Б)

1.12. AB = (3 - 3; - 4 - (-1)) = (0; -3)
|AB| = sqrt(0^2 + (-3)^2) = sqrt(9) = 3.
Ответ: В)

2.1.
$( \frac{x - 2y}{ {x}^{2} + 2xy } - \frac{x + 2y}{ {x}^{2} - 2xy } ) \div \frac{4 {y}^{2} }{4 {y}^{2} - {x}^{2} } = ( \frac{x - 2y}{x(x + 2y)} - \frac{x + 2y}{x(x - 2y)} ) \div \frac{4 {y}^{2} }{(2y - x)(2y + x)} = ( \frac{ {(x - 2y)}^{2} }{x(x + 2y)(x - 2y)} - \frac{ {(x + 2y)}^{2} }{x(x - 2y)(x + 2y)} ) \div \frac{4 {y}^{2} }{(2y - x)(2y + x)} = \frac{(x - 2y + x + 2y)(x - 2y - x - 2y)}{x(x - 2y)(x + 2y)} \times \frac{(2y - x)(2y + x)}{4 {y}^{2} } = \frac{2x \times ( - 4y)}{x(x - 2y)(x + 2y)} \times \frac{ - (x - 2y)(x + 2y)}{4 {y}^{2} } = \frac {8xy}{4x {y}^{2}}=\frac{2}{y}$